How to find the name of the variables used in a script in bash - linux

I am trying to list the names of the variables that I have declared in my script, so I have been trying with different combinations of:
( set -o posix ; set ) | less
But I am not getting the result that I expected. It gives me an ouput with all variables in addition to my script´s variables. I do not know if I am using this command in the correct way. My intention is to only list the variables´ names that I have declared in my script and get an output like this:
VAR1
VAR2
VAR3
VAR4
…

set cannot be told to list only variables declared in a script. What can be done is to temporarily store defined variables at the script's beginning and filter them out at the end, so only variables defined in-between remain:
#!/bin/bash
set -o posix
set|sed s/=.\*// >/tmp/$$
VAR4=4
VAR3=3
VAR2=2
VAR1=1
set|sed s/=.\*//|grep -vf/tmp/$$
rm /tmp/$$
Note Socowi's valuable suggestions for improvement below.

A very simple solution as long as you use the prefix for your own variables. You should do this to avoid accidentally overwriting existing variables. Set defaults returns to a very long list of all defined variables not just yours, so you need to filter it by prefix.
#!/bin/bash
my_var1=aaa
my_var2=`ls`
export my_var3=bbb
compgen -v| awk -F= '/^my_/{print $1}'
# If you use sh shell it remains
#set |awk -F= '/^my_/{print $1}'
result
my_var1
my_var2
my_var3
EDIT: #Socowi rightly suggested using compgen -v for bash instead of set. Compgen is built into bash. Thanks.

Related

use an env variable as part of the name of a different env variable in bash [duplicate]

I am confused about a bash script.
I have the following code:
function grep_search() {
magic_way_to_define_magic_variable_$1=`ls | tail -1`
echo $magic_variable_$1
}
I want to be able to create a variable name containing the first argument of the command and bearing the value of e.g. the last line of ls.
So to illustrate what I want:
$ ls | tail -1
stack-overflow.txt
$ grep_search() open_box
stack-overflow.txt
So, how should I define/declare $magic_way_to_define_magic_variable_$1 and how should I call it within the script?
I have tried eval, ${...}, \$${...}, but I am still confused.
I've been looking for better way of doing it recently. Associative array sounded like overkill for me. Look what I found:
suffix=bzz
declare prefix_$suffix=mystr
...and then...
varname=prefix_$suffix
echo ${!varname}
From the docs:
The ‘$’ character introduces parameter expansion, command substitution, or arithmetic expansion. ...
The basic form of parameter expansion is ${parameter}. The value of parameter is substituted. ...
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion. The value is subject to tilde expansion, parameter expansion, command substitution, and arithmetic expansion. ...
Use an associative array, with command names as keys.
# Requires bash 4, though
declare -A magic_variable=()
function grep_search() {
magic_variable[$1]=$( ls | tail -1 )
echo ${magic_variable[$1]}
}
If you can't use associative arrays (e.g., you must support bash 3), you can use declare to create dynamic variable names:
declare "magic_variable_$1=$(ls | tail -1)"
and use indirect parameter expansion to access the value.
var="magic_variable_$1"
echo "${!var}"
See BashFAQ: Indirection - Evaluating indirect/reference variables.
Beyond associative arrays, there are several ways of achieving dynamic variables in Bash. Note that all these techniques present risks, which are discussed at the end of this answer.
In the following examples I will assume that i=37 and that you want to alias the variable named var_37 whose initial value is lolilol.
Method 1. Using a “pointer” variable
You can simply store the name of the variable in an indirection variable, not unlike a C pointer. Bash then has a syntax for reading the aliased variable: ${!name} expands to the value of the variable whose name is the value of the variable name. You can think of it as a two-stage expansion: ${!name} expands to $var_37, which expands to lolilol.
name="var_$i"
echo "$name" # outputs “var_37”
echo "${!name}" # outputs “lolilol”
echo "${!name%lol}" # outputs “loli”
# etc.
Unfortunately, there is no counterpart syntax for modifying the aliased variable. Instead, you can achieve assignment with one of the following tricks.
1a. Assigning with eval
eval is evil, but is also the simplest and most portable way of achieving our goal. You have to carefully escape the right-hand side of the assignment, as it will be evaluated twice. An easy and systematic way of doing this is to evaluate the right-hand side beforehand (or to use printf %q).
And you should check manually that the left-hand side is a valid variable name, or a name with index (what if it was evil_code # ?). By contrast, all other methods below enforce it automatically.
# check that name is a valid variable name:
# note: this code does not support variable_name[index]
shopt -s globasciiranges
[[ "$name" == [a-zA-Z_]*([a-zA-Z_0-9]) ]] || exit
value='babibab'
eval "$name"='$value' # carefully escape the right-hand side!
echo "$var_37" # outputs “babibab”
Downsides:
does not check the validity of the variable name.
eval is evil.
eval is evil.
eval is evil.
1b. Assigning with read
The read builtin lets you assign values to a variable of which you give the name, a fact which can be exploited in conjunction with here-strings:
IFS= read -r -d '' "$name" <<< 'babibab'
echo "$var_37" # outputs “babibab\n”
The IFS part and the option -r make sure that the value is assigned as-is, while the option -d '' allows to assign multi-line values. Because of this last option, the command returns with an non-zero exit code.
Note that, since we are using a here-string, a newline character is appended to the value.
Downsides:
somewhat obscure;
returns with a non-zero exit code;
appends a newline to the value.
1c. Assigning with printf
Since Bash 3.1 (released 2005), the printf builtin can also assign its result to a variable whose name is given. By contrast with the previous solutions, it just works, no extra effort is needed to escape things, to prevent splitting and so on.
printf -v "$name" '%s' 'babibab'
echo "$var_37" # outputs “babibab”
Downsides:
Less portable (but, well).
Method 2. Using a “reference” variable
Since Bash 4.3 (released 2014), the declare builtin has an option -n for creating a variable which is a “name reference” to another variable, much like C++ references. Just as in Method 1, the reference stores the name of the aliased variable, but each time the reference is accessed (either for reading or assigning), Bash automatically resolves the indirection.
In addition, Bash has a special and very confusing syntax for getting the value of the reference itself, judge by yourself: ${!ref}.
declare -n ref="var_$i"
echo "${!ref}" # outputs “var_37”
echo "$ref" # outputs “lolilol”
ref='babibab'
echo "$var_37" # outputs “babibab”
This does not avoid the pitfalls explained below, but at least it makes the syntax straightforward.
Downsides:
Not portable.
Risks
All these aliasing techniques present several risks. The first one is executing arbitrary code each time you resolve the indirection (either for reading or for assigning). Indeed, instead of a scalar variable name, like var_37, you may as well alias an array subscript, like arr[42]. But Bash evaluates the contents of the square brackets each time it is needed, so aliasing arr[$(do_evil)] will have unexpected effects… As a consequence, only use these techniques when you control the provenance of the alias.
function guillemots {
declare -n var="$1"
var="«${var}»"
}
arr=( aaa bbb ccc )
guillemots 'arr[1]' # modifies the second cell of the array, as expected
guillemots 'arr[$(date>>date.out)1]' # writes twice into date.out
# (once when expanding var, once when assigning to it)
The second risk is creating a cyclic alias. As Bash variables are identified by their name and not by their scope, you may inadvertently create an alias to itself (while thinking it would alias a variable from an enclosing scope). This may happen in particular when using common variable names (like var). As a consequence, only use these techniques when you control the name of the aliased variable.
function guillemots {
# var is intended to be local to the function,
# aliasing a variable which comes from outside
declare -n var="$1"
var="«${var}»"
}
var='lolilol'
guillemots var # Bash warnings: “var: circular name reference”
echo "$var" # outputs anything!
Source:
BashFaq/006: How can I use variable variables (indirect variables, pointers, references) or associative arrays?
BashFAQ/048: eval command and security issues
Example below returns value of $name_of_var
var=name_of_var
echo $(eval echo "\$$var")
Use declare
There is no need on using prefixes like on other answers, neither arrays. Use just declare, double quotes, and parameter expansion.
I often use the following trick to parse argument lists contanining one to n arguments formatted as key=value otherkey=othervalue etc=etc, Like:
# brace expansion just to exemplify
for variable in {one=foo,two=bar,ninja=tip}
do
declare "${variable%=*}=${variable#*=}"
done
echo $one $two $ninja
# foo bar tip
But expanding the argv list like
for v in "$#"; do declare "${v%=*}=${v#*=}"; done
Extra tips
# parse argv's leading key=value parameters
for v in "$#"; do
case "$v" in ?*=?*) declare "${v%=*}=${v#*=}";; *) break;; esac
done
# consume argv's leading key=value parameters
while test $# -gt 0; do
case "$1" in ?*=?*) declare "${1%=*}=${1#*=}";; *) break;; esac
shift
done
Combining two highly rated answers here into a complete example that is hopefully useful and self-explanatory:
#!/bin/bash
intro="You know what,"
pet1="cat"
pet2="chicken"
pet3="cow"
pet4="dog"
pet5="pig"
# Setting and reading dynamic variables
for i in {1..5}; do
pet="pet$i"
declare "sentence$i=$intro I have a pet ${!pet} at home"
done
# Just reading dynamic variables
for i in {1..5}; do
sentence="sentence$i"
echo "${!sentence}"
done
echo
echo "Again, but reading regular variables:"
echo $sentence1
echo $sentence2
echo $sentence3
echo $sentence4
echo $sentence5
Output:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
Again, but reading regular variables:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
This will work too
my_country_code="green"
x="country"
eval z='$'my_"$x"_code
echo $z ## o/p: green
In your case
eval final_val='$'magic_way_to_define_magic_variable_"$1"
echo $final_val
This should work:
function grep_search() {
declare magic_variable_$1="$(ls | tail -1)"
echo "$(tmpvar=magic_variable_$1 && echo ${!tmpvar})"
}
grep_search var # calling grep_search with argument "var"
An extra method that doesn't rely on which shell/bash version you have is by using envsubst. For example:
newvar=$(echo '$magic_variable_'"${dynamic_part}" | envsubst)
For zsh (newers mac os versions), you should use
real_var="holaaaa"
aux_var="real_var"
echo ${(P)aux_var}
holaaaa
Instead of "!"
As per BashFAQ/006, you can use read with here string syntax for assigning indirect variables:
function grep_search() {
read "$1" <<<$(ls | tail -1);
}
Usage:
$ grep_search open_box
$ echo $open_box
stack-overflow.txt
Even though it's an old question, I still had some hard time with fetching dynamic variables names, while avoiding the eval (evil) command.
Solved it with declare -n which creates a reference to a dynamic value, this is especially useful in CI/CD processes, where the required secret names of the CI/CD service are not known until runtime. Here's how:
# Bash v4.3+
# -----------------------------------------------------------
# Secerts in CI/CD service, injected as environment variables
# AWS_ACCESS_KEY_ID_DEV, AWS_SECRET_ACCESS_KEY_DEV
# AWS_ACCESS_KEY_ID_STG, AWS_SECRET_ACCESS_KEY_STG
# -----------------------------------------------------------
# Environment variables injected by CI/CD service
# BRANCH_NAME="DEV"
# -----------------------------------------------------------
declare -n _AWS_ACCESS_KEY_ID_REF=AWS_ACCESS_KEY_ID_${BRANCH_NAME}
declare -n _AWS_SECRET_ACCESS_KEY_REF=AWS_SECRET_ACCESS_KEY_${BRANCH_NAME}
export AWS_ACCESS_KEY_ID=${_AWS_ACCESS_KEY_ID_REF}
export AWS_SECRET_ACCESS_KEY=${_AWS_SECRET_ACCESS_KEY_REF}
echo $AWS_ACCESS_KEY_ID $AWS_SECRET_ACCESS_KEY
aws s3 ls
Wow, most of the syntax is horrible! Here is one solution with some simpler syntax if you need to indirectly reference arrays:
#!/bin/bash
foo_1=(fff ddd) ;
foo_2=(ggg ccc) ;
for i in 1 2 ;
do
eval mine=( \${foo_$i[#]} ) ;
echo ${mine[#]}" " ;
done ;
For simpler use cases I recommend the syntax described in the Advanced Bash-Scripting Guide.
KISS approach:
a=1
c="bam"
let "$c$a"=4
echo $bam1
results in 4
I want to be able to create a variable name containing the first argument of the command
script.sh file:
#!/usr/bin/env bash
function grep_search() {
eval $1=$(ls | tail -1)
}
Test:
$ source script.sh
$ grep_search open_box
$ echo $open_box
script.sh
As per help eval:
Execute arguments as a shell command.
You may also use Bash ${!var} indirect expansion, as already mentioned, however it doesn't support retrieving of array indices.
For further read or examples, check BashFAQ/006 about Indirection.
We are not aware of any trick that can duplicate that functionality in POSIX or Bourne shells without eval, which can be difficult to do securely. So, consider this a use at your own risk hack.
However, you should re-consider using indirection as per the following notes.
Normally, in bash scripting, you won't need indirect references at all. Generally, people look at this for a solution when they don't understand or know about Bash Arrays or haven't fully considered other Bash features such as functions.
Putting variable names or any other bash syntax inside parameters is frequently done incorrectly and in inappropriate situations to solve problems that have better solutions. It violates the separation between code and data, and as such puts you on a slippery slope toward bugs and security issues. Indirection can make your code less transparent and harder to follow.
For indexed arrays, you can reference them like so:
foo=(a b c)
bar=(d e f)
for arr_var in 'foo' 'bar'; do
declare -a 'arr=("${'"$arr_var"'[#]}")'
# do something with $arr
echo "\$$arr_var contains:"
for char in "${arr[#]}"; do
echo "$char"
done
done
Associative arrays can be referenced similarly but need the -A switch on declare instead of -a.
POSIX compliant answer
For this solution you'll need to have r/w permissions to the /tmp folder.
We create a temporary file holding our variables and leverage the -a flag of the set built-in:
$ man set
...
-a Each variable or function that is created or modified is given the export attribute and marked for export to the environment of subsequent commands.
Therefore, if we create a file holding our dynamic variables, we can use set to bring them to life inside our script.
The implementation
#!/bin/sh
# Give the temp file a unique name so you don't mess with any other files in there
ENV_FILE="/tmp/$(date +%s)"
MY_KEY=foo
MY_VALUE=bar
echo "$MY_KEY=$MY_VALUE" >> "$ENV_FILE"
# Now that our env file is created and populated, we can use "set"
set -a; . "$ENV_FILE"; set +a
rm "$ENV_FILE"
echo "$foo"
# Output is "bar" (without quotes)
Explaining the steps above:
# Enables the -a behavior
set -a
# Sources the env file
. "$ENV_FILE"
# Disables the -a behavior
set +a
While I think declare -n is still the best way to do it there is another way nobody mentioned it, very useful in CI/CD
function dynamic(){
export a_$1="bla"
}
dynamic 2
echo $a_2
This function will not support spaces so dynamic "2 3" will return an error.
for varname=$prefix_suffix format, just use:
varname=${prefix}_suffix

How to define and print a variable value at the same time in shell

I am facing issue in one bash script as I have print the value of a variable and after then using that variable as an input to another variable but values are not getting print as I understand I have assign value to a variable and then can print.
So I want to know can I define and print a variable value at the same time and store the value to use in another variable within the scope of the function.
For ex.
a="My name is Vikas"
echo "a=My name is Vikas"
b="${a} Singh"
echo "b=${a} Singh"
but I want to do the below
echo "a=My name is Vikas"
echo "b=${a} Singh"
so output should be like
a=My name is Vikas
b=My name is Vikas Singh.
Please help
an assignment can be done while doing an expansion in the case variables are not already set
a= b=
echo "a=${a:=My name is Vikas}"
echo "b=${b:=${a} Singh}"
from manual
${parameter:=word}
If parameter is unset or null, the expansion of word is assigned to parameter. The value of parameter is then substituted. Positional parameters and special parameters may not be assigned to in this way.
a time later, I realized that this could be misleading : it is echo "a=.." which is doing assignment but expansion, which made me understand the question differently, assignments (like any other commands) can be traced with set -x or set -o xtrace (to undo set +x or set +o xtrace) (another similar option set -v or set -o verbose show the command before expansion)
set -x
a="My name is Vikas"
b="${a} Singh"
set +x
will write assignments after expansion to (file descriptor 2) standard error.
You can certainly print the contents of a variable into another variable. The command to print is printf, and the option -v prints to a variable. From the man page:
printf [-v var] format [arguments]
Write the formatted arguments to the standard output under the control of
the format. The -v option causes the output to be assigned to the vari-
able var rather than being printed to the standard output.
So...
$ fname="Vikas"; lname="Singh"
$ printf -v a 'My name is %s' "$fname"
$ printf -v b '%s %s' "$a" "$lname"
$ printf '%s\n%s\n' "$a" "$b"
My name is Vikas
My name is Vikas Singh
But if you want your output to include the variable names, you should probably use declare -p:
$ declare -p a b
declare -- a="My name is Vikas"
declare -- b="My name is Vikas Singh"
By using declare -p, your output may be recycled as input to recreate the same variables in the future.
I think you were trying to do something like this.
b="${a='My name is Vikas'} Singh"
echo $a
echo $b
The command within curly braces { and } sets the variable a in the current shell, and the outer assigment sets b.
Bash manual
EDIT: Actually, as Nauhel pointed out, the variables should not be set. So it should begin with unset a.

2-part String to variable names bash [duplicate]

I am confused about a bash script.
I have the following code:
function grep_search() {
magic_way_to_define_magic_variable_$1=`ls | tail -1`
echo $magic_variable_$1
}
I want to be able to create a variable name containing the first argument of the command and bearing the value of e.g. the last line of ls.
So to illustrate what I want:
$ ls | tail -1
stack-overflow.txt
$ grep_search() open_box
stack-overflow.txt
So, how should I define/declare $magic_way_to_define_magic_variable_$1 and how should I call it within the script?
I have tried eval, ${...}, \$${...}, but I am still confused.
I've been looking for better way of doing it recently. Associative array sounded like overkill for me. Look what I found:
suffix=bzz
declare prefix_$suffix=mystr
...and then...
varname=prefix_$suffix
echo ${!varname}
From the docs:
The ‘$’ character introduces parameter expansion, command substitution, or arithmetic expansion. ...
The basic form of parameter expansion is ${parameter}. The value of parameter is substituted. ...
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion. The value is subject to tilde expansion, parameter expansion, command substitution, and arithmetic expansion. ...
Use an associative array, with command names as keys.
# Requires bash 4, though
declare -A magic_variable=()
function grep_search() {
magic_variable[$1]=$( ls | tail -1 )
echo ${magic_variable[$1]}
}
If you can't use associative arrays (e.g., you must support bash 3), you can use declare to create dynamic variable names:
declare "magic_variable_$1=$(ls | tail -1)"
and use indirect parameter expansion to access the value.
var="magic_variable_$1"
echo "${!var}"
See BashFAQ: Indirection - Evaluating indirect/reference variables.
Beyond associative arrays, there are several ways of achieving dynamic variables in Bash. Note that all these techniques present risks, which are discussed at the end of this answer.
In the following examples I will assume that i=37 and that you want to alias the variable named var_37 whose initial value is lolilol.
Method 1. Using a “pointer” variable
You can simply store the name of the variable in an indirection variable, not unlike a C pointer. Bash then has a syntax for reading the aliased variable: ${!name} expands to the value of the variable whose name is the value of the variable name. You can think of it as a two-stage expansion: ${!name} expands to $var_37, which expands to lolilol.
name="var_$i"
echo "$name" # outputs “var_37”
echo "${!name}" # outputs “lolilol”
echo "${!name%lol}" # outputs “loli”
# etc.
Unfortunately, there is no counterpart syntax for modifying the aliased variable. Instead, you can achieve assignment with one of the following tricks.
1a. Assigning with eval
eval is evil, but is also the simplest and most portable way of achieving our goal. You have to carefully escape the right-hand side of the assignment, as it will be evaluated twice. An easy and systematic way of doing this is to evaluate the right-hand side beforehand (or to use printf %q).
And you should check manually that the left-hand side is a valid variable name, or a name with index (what if it was evil_code # ?). By contrast, all other methods below enforce it automatically.
# check that name is a valid variable name:
# note: this code does not support variable_name[index]
shopt -s globasciiranges
[[ "$name" == [a-zA-Z_]*([a-zA-Z_0-9]) ]] || exit
value='babibab'
eval "$name"='$value' # carefully escape the right-hand side!
echo "$var_37" # outputs “babibab”
Downsides:
does not check the validity of the variable name.
eval is evil.
eval is evil.
eval is evil.
1b. Assigning with read
The read builtin lets you assign values to a variable of which you give the name, a fact which can be exploited in conjunction with here-strings:
IFS= read -r -d '' "$name" <<< 'babibab'
echo "$var_37" # outputs “babibab\n”
The IFS part and the option -r make sure that the value is assigned as-is, while the option -d '' allows to assign multi-line values. Because of this last option, the command returns with an non-zero exit code.
Note that, since we are using a here-string, a newline character is appended to the value.
Downsides:
somewhat obscure;
returns with a non-zero exit code;
appends a newline to the value.
1c. Assigning with printf
Since Bash 3.1 (released 2005), the printf builtin can also assign its result to a variable whose name is given. By contrast with the previous solutions, it just works, no extra effort is needed to escape things, to prevent splitting and so on.
printf -v "$name" '%s' 'babibab'
echo "$var_37" # outputs “babibab”
Downsides:
Less portable (but, well).
Method 2. Using a “reference” variable
Since Bash 4.3 (released 2014), the declare builtin has an option -n for creating a variable which is a “name reference” to another variable, much like C++ references. Just as in Method 1, the reference stores the name of the aliased variable, but each time the reference is accessed (either for reading or assigning), Bash automatically resolves the indirection.
In addition, Bash has a special and very confusing syntax for getting the value of the reference itself, judge by yourself: ${!ref}.
declare -n ref="var_$i"
echo "${!ref}" # outputs “var_37”
echo "$ref" # outputs “lolilol”
ref='babibab'
echo "$var_37" # outputs “babibab”
This does not avoid the pitfalls explained below, but at least it makes the syntax straightforward.
Downsides:
Not portable.
Risks
All these aliasing techniques present several risks. The first one is executing arbitrary code each time you resolve the indirection (either for reading or for assigning). Indeed, instead of a scalar variable name, like var_37, you may as well alias an array subscript, like arr[42]. But Bash evaluates the contents of the square brackets each time it is needed, so aliasing arr[$(do_evil)] will have unexpected effects… As a consequence, only use these techniques when you control the provenance of the alias.
function guillemots {
declare -n var="$1"
var="«${var}»"
}
arr=( aaa bbb ccc )
guillemots 'arr[1]' # modifies the second cell of the array, as expected
guillemots 'arr[$(date>>date.out)1]' # writes twice into date.out
# (once when expanding var, once when assigning to it)
The second risk is creating a cyclic alias. As Bash variables are identified by their name and not by their scope, you may inadvertently create an alias to itself (while thinking it would alias a variable from an enclosing scope). This may happen in particular when using common variable names (like var). As a consequence, only use these techniques when you control the name of the aliased variable.
function guillemots {
# var is intended to be local to the function,
# aliasing a variable which comes from outside
declare -n var="$1"
var="«${var}»"
}
var='lolilol'
guillemots var # Bash warnings: “var: circular name reference”
echo "$var" # outputs anything!
Source:
BashFaq/006: How can I use variable variables (indirect variables, pointers, references) or associative arrays?
BashFAQ/048: eval command and security issues
Example below returns value of $name_of_var
var=name_of_var
echo $(eval echo "\$$var")
Use declare
There is no need on using prefixes like on other answers, neither arrays. Use just declare, double quotes, and parameter expansion.
I often use the following trick to parse argument lists contanining one to n arguments formatted as key=value otherkey=othervalue etc=etc, Like:
# brace expansion just to exemplify
for variable in {one=foo,two=bar,ninja=tip}
do
declare "${variable%=*}=${variable#*=}"
done
echo $one $two $ninja
# foo bar tip
But expanding the argv list like
for v in "$#"; do declare "${v%=*}=${v#*=}"; done
Extra tips
# parse argv's leading key=value parameters
for v in "$#"; do
case "$v" in ?*=?*) declare "${v%=*}=${v#*=}";; *) break;; esac
done
# consume argv's leading key=value parameters
while test $# -gt 0; do
case "$1" in ?*=?*) declare "${1%=*}=${1#*=}";; *) break;; esac
shift
done
Combining two highly rated answers here into a complete example that is hopefully useful and self-explanatory:
#!/bin/bash
intro="You know what,"
pet1="cat"
pet2="chicken"
pet3="cow"
pet4="dog"
pet5="pig"
# Setting and reading dynamic variables
for i in {1..5}; do
pet="pet$i"
declare "sentence$i=$intro I have a pet ${!pet} at home"
done
# Just reading dynamic variables
for i in {1..5}; do
sentence="sentence$i"
echo "${!sentence}"
done
echo
echo "Again, but reading regular variables:"
echo $sentence1
echo $sentence2
echo $sentence3
echo $sentence4
echo $sentence5
Output:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
Again, but reading regular variables:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
This will work too
my_country_code="green"
x="country"
eval z='$'my_"$x"_code
echo $z ## o/p: green
In your case
eval final_val='$'magic_way_to_define_magic_variable_"$1"
echo $final_val
This should work:
function grep_search() {
declare magic_variable_$1="$(ls | tail -1)"
echo "$(tmpvar=magic_variable_$1 && echo ${!tmpvar})"
}
grep_search var # calling grep_search with argument "var"
An extra method that doesn't rely on which shell/bash version you have is by using envsubst. For example:
newvar=$(echo '$magic_variable_'"${dynamic_part}" | envsubst)
For zsh (newers mac os versions), you should use
real_var="holaaaa"
aux_var="real_var"
echo ${(P)aux_var}
holaaaa
Instead of "!"
As per BashFAQ/006, you can use read with here string syntax for assigning indirect variables:
function grep_search() {
read "$1" <<<$(ls | tail -1);
}
Usage:
$ grep_search open_box
$ echo $open_box
stack-overflow.txt
Even though it's an old question, I still had some hard time with fetching dynamic variables names, while avoiding the eval (evil) command.
Solved it with declare -n which creates a reference to a dynamic value, this is especially useful in CI/CD processes, where the required secret names of the CI/CD service are not known until runtime. Here's how:
# Bash v4.3+
# -----------------------------------------------------------
# Secerts in CI/CD service, injected as environment variables
# AWS_ACCESS_KEY_ID_DEV, AWS_SECRET_ACCESS_KEY_DEV
# AWS_ACCESS_KEY_ID_STG, AWS_SECRET_ACCESS_KEY_STG
# -----------------------------------------------------------
# Environment variables injected by CI/CD service
# BRANCH_NAME="DEV"
# -----------------------------------------------------------
declare -n _AWS_ACCESS_KEY_ID_REF=AWS_ACCESS_KEY_ID_${BRANCH_NAME}
declare -n _AWS_SECRET_ACCESS_KEY_REF=AWS_SECRET_ACCESS_KEY_${BRANCH_NAME}
export AWS_ACCESS_KEY_ID=${_AWS_ACCESS_KEY_ID_REF}
export AWS_SECRET_ACCESS_KEY=${_AWS_SECRET_ACCESS_KEY_REF}
echo $AWS_ACCESS_KEY_ID $AWS_SECRET_ACCESS_KEY
aws s3 ls
Wow, most of the syntax is horrible! Here is one solution with some simpler syntax if you need to indirectly reference arrays:
#!/bin/bash
foo_1=(fff ddd) ;
foo_2=(ggg ccc) ;
for i in 1 2 ;
do
eval mine=( \${foo_$i[#]} ) ;
echo ${mine[#]}" " ;
done ;
For simpler use cases I recommend the syntax described in the Advanced Bash-Scripting Guide.
KISS approach:
a=1
c="bam"
let "$c$a"=4
echo $bam1
results in 4
I want to be able to create a variable name containing the first argument of the command
script.sh file:
#!/usr/bin/env bash
function grep_search() {
eval $1=$(ls | tail -1)
}
Test:
$ source script.sh
$ grep_search open_box
$ echo $open_box
script.sh
As per help eval:
Execute arguments as a shell command.
You may also use Bash ${!var} indirect expansion, as already mentioned, however it doesn't support retrieving of array indices.
For further read or examples, check BashFAQ/006 about Indirection.
We are not aware of any trick that can duplicate that functionality in POSIX or Bourne shells without eval, which can be difficult to do securely. So, consider this a use at your own risk hack.
However, you should re-consider using indirection as per the following notes.
Normally, in bash scripting, you won't need indirect references at all. Generally, people look at this for a solution when they don't understand or know about Bash Arrays or haven't fully considered other Bash features such as functions.
Putting variable names or any other bash syntax inside parameters is frequently done incorrectly and in inappropriate situations to solve problems that have better solutions. It violates the separation between code and data, and as such puts you on a slippery slope toward bugs and security issues. Indirection can make your code less transparent and harder to follow.
For indexed arrays, you can reference them like so:
foo=(a b c)
bar=(d e f)
for arr_var in 'foo' 'bar'; do
declare -a 'arr=("${'"$arr_var"'[#]}")'
# do something with $arr
echo "\$$arr_var contains:"
for char in "${arr[#]}"; do
echo "$char"
done
done
Associative arrays can be referenced similarly but need the -A switch on declare instead of -a.
POSIX compliant answer
For this solution you'll need to have r/w permissions to the /tmp folder.
We create a temporary file holding our variables and leverage the -a flag of the set built-in:
$ man set
...
-a Each variable or function that is created or modified is given the export attribute and marked for export to the environment of subsequent commands.
Therefore, if we create a file holding our dynamic variables, we can use set to bring them to life inside our script.
The implementation
#!/bin/sh
# Give the temp file a unique name so you don't mess with any other files in there
ENV_FILE="/tmp/$(date +%s)"
MY_KEY=foo
MY_VALUE=bar
echo "$MY_KEY=$MY_VALUE" >> "$ENV_FILE"
# Now that our env file is created and populated, we can use "set"
set -a; . "$ENV_FILE"; set +a
rm "$ENV_FILE"
echo "$foo"
# Output is "bar" (without quotes)
Explaining the steps above:
# Enables the -a behavior
set -a
# Sources the env file
. "$ENV_FILE"
# Disables the -a behavior
set +a
While I think declare -n is still the best way to do it there is another way nobody mentioned it, very useful in CI/CD
function dynamic(){
export a_$1="bla"
}
dynamic 2
echo $a_2
This function will not support spaces so dynamic "2 3" will return an error.
for varname=$prefix_suffix format, just use:
varname=${prefix}_suffix

Extract all variable values in a shell script

I'm debugging an old shell script; I want to check the values of all the variables used, it's a huge ugly script with approx more than 140 variables used. Is there anyway I can extract the variable names from the script and put them in a convenient pattern like:
#!/bin/sh
if [ ${BLAH} ....
.....
rm -rf ${JUNK}.....
to
echo ${BLAH}
echo ${JUNK}
...
Try running your script as follows:
bash -x ./script.bash
Or enable the setting in the script:
set -x
You can dump all interested variables in one command using:
set | grep -w -e BLAH -e JUNK
To dump all the variables to stdout use:
set
or
env
from inside your script.
You can extract a (sub)list of the variables declared in your script using grep:
grep -Po "([a-z][a-zA-Z0-9_]+)(?==\")" ./script.bash | sort -u
Disclaimer: why "sublist"?
The expression given will match string followed by an egal sign (=) and a double quote ("). So if you don't use syntax such as myvar="my-value" it won't work.
But you got the idea.
grep Options
-P --perl-regexp: Interpret PATTERN as a Perl regular expression (PCRE, see below) (experimental) ;
-o --only-matching: Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Pattern
I'm using a positive lookahead: (?==\") to require an egal sign followed by a double quote.
In bash, but not sh, compgen -v will list the names of all variables assigned (compare this to set, which has a great deal of output other than variable names, and thus needs to be parsed).
Thus, if you change the top of the script to #!/bin/bash, you will be able to use compgen -v to generate that list.
That said, the person who advised you use set -x did well. Consider this extension on that:
PS4=':$BASH_SOURCE:$LINENO+'; set -x
This will print the source file and line number before every command (or variable assignment) which is executed, so you will have a log not only of which variables are set, but just where in the source each one was assigned. This makes tracking down where each variable is set far easier.

How to substitute shell variables in complex text files

I have several text files in which I have introduced shell variables ($VAR1 or $VAR2 for instance).
I would like to take those files (one by one) and save them in new files where all variables would have been replaced.
To do this, I used the following shell script (found on StackOverflow):
while read line
do
eval echo "$line" >> destination.txt
done < "source.txt"
This works very well on very basic files.
But on more complex files, the "eval" command does too much:
Lines starting with "#" are skipped
XML files parsing results in tons of errors
Is there a better way to do it? (in shell script... I know this is easily done with Ant for instance)
Kind regards
Looking, it turns out on my system there is an envsubst command which is part of the gettext-base package.
So, this makes it easy:
envsubst < "source.txt" > "destination.txt"
Note if you want to use the same file for both, you'll have to use something like moreutil's sponge, as suggested by Johnny Utahh: envsubst < "source.txt" | sponge "source.txt". (Because the shell redirect will otherwise empty the file before its read.)
In reference to answer 2, when discussing envsubst, you asked:
How can I make it work with the variables that are declared in my .sh script?
The answer is you simply need to export your variables before calling envsubst.
You can also limit the variable strings you want to replace in the input using the envsubst SHELL_FORMAT argument (avoiding the unintended replacement of a string in the input with a common shell variable value - e.g. $HOME).
For instance:
export VAR1='somevalue' VAR2='someothervalue'
MYVARS='$VAR1:$VAR2'
envsubst "$MYVARS" <source.txt >destination.txt
Will replace all instances of $VAR1 and $VAR2 (and only VAR1 and VAR2) in source.txt with 'somevalue' and 'someothervalue' respectively.
I know this topic is old, but I have a simpler working solution without exporting the variables. Can be a oneliner, but I prefer to split using \ on line end.
var1='myVar1'\
var2=2\
var3=${var1}\
envsubst '$var1,$var3' < "source.txt" > "destination.txt"
# ^^^^^^^^^^^ ^^^^^^^^^^ ^^^^^^^^^^^^^^^
# define which to replace input output
The variables need to be defined to the same line as envsubst is to get considered as environment variables.
The '$var1,$var3' is optional to only replace the specified ones. Imagine an input file containing ${VARIABLE_USED_BY_JENKINS} which should not be replaced.
Define your ENV variable
$ export MY_ENV_VAR=congratulation
Create template file (in.txt) with following content
$MY_ENV_VAR
You can also use all other ENV variables defined by your system like (in linux) $TERM, $SHELL, $HOME...
Run this command to raplace all env-variables in your in.txt file and to write the result to out.txt
$ envsubst "`printf '${%s} ' $(sh -c "env|cut -d'=' -f1")`" < in.txt > out.txt
Check the content of out.txt file
$ cat out.txt
and you should see "congratulation".
There is also this option:
define your variables in a file
$ cat variables.env
# info about what this var is
export var1=a
# info about var again
export var2=b
define a template file that uses the variables
$ cat file1-template.txt
This is var1: "${var1}"
This is var2: "${var2}"
generate the final file, with variables replaced with values
$ source variables.env
$ envsubst < file1-template.txt > file1.txt
$ cat file1.txt
This is var1: "a"
This is var2: "b"
If you want env variables to be replaced in your source files while keeping all of the non env variables as they are, you can use the following command:
envsubst "$(printf '${%s} ' $(env | sed 's/=.*//'))" < source.txt > destination.txt
The syntax for replacing only specific variables is explained here. The command above is using a sub-shell to list all defined variables and then passing it to the envsubst
So if there's a defined env variable called $NAME, and your source.txt file looks like this:
Hello $NAME
Your balance is 123 ($USD)
The destination.txt will be:
Hello Arik
Your balance is 123 ($USD)
Notice that the $NAME is replaced and the $USD is left untouched
while IFS='=' read -r name value ; do
# Print line if found variable
sed -n '/${'"${name}"'}/p' docker-compose.yml
# Replace variable with value.
sed -i 's|${'"${name}"'}|'"${value}"'|' docker-compose.yml
done < <(env)
Note: Variable name or value should not contain "|", because it is used as a delimiter.
If you really only want to use bash (and sed), then I would go through each of your environment variables (as returned by set in posix mode) and build a bunch of -e 'regex' for sed from that, terminated by a -e 's/\$[a-zA-Z_][a-zA-Z0-9_]*//g', then pass all that to sed.
Perl would do a nicer job though, you have access to the environment vars as an array and you can do executable replacements so you only match any environment variable once.
Actually you need to change your read to read -r which will make it ignore backslashes.
Also, you should escape quotes and backslashes.
So
while read -r line; do
line="${line//\\/\\\\}"
line="${line//\"/\\\"}"
line="${line//\`/\\\`}"
eval echo "\"$line\""
done > destination.txt < source.txt
Still a terrible way to do expansion though.
Export all the needed variables and then use a perl onliner
TEXT=$(echo "$TEXT"|perl -wpne 's#\${?(\w+)}?# $ENV{$1} // $& #ge;')
This will replace all the ENV variables present in TEXT with actual values.
Quotes are also preserved :)
Call the perl binary, in search and replace per line mode ( the -pi ) by running the perl code ( the -e) in the single quotes, which iterates over the keys of the special %ENV hash containing the exported variable names as keys and the exported variable values as the keys' values and for each iteration simple replace a string containing a $<<key>> with its <<value>>.
perl -pi -e 'foreach $key(sort keys %ENV){ s/\$$key/$ENV{$key}/g}' file
Caveat:
An additional logic handling is required for cases in which two or more vars start with the same string ...
envsubst seems exactly like something I wanted to use, but -v option surprised me a bit.
While envsubst < template.txt was working fine, the same with option -v was not working:
$ cat /etc/redhat-release
Red Hat Enterprise Linux Server release 7.1 (Maipo)
$ envsubst -V
envsubst (GNU gettext-runtime) 0.18.2
Copyright (C) 2003-2007 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
Written by Bruno Haible.
As I wrote, this was not working:
$ envsubst -v < template.txt
envsubst: missing arguments
$ cat template.txt | envsubst -v
envsubst: missing arguments
I had to do this to make it work:
TEXT=`cat template.txt`; envsubst -v "$TEXT"
Maybe it helps someone.

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