Let's suppose I have following two variables:
arg1=5
count5="test"
Now, I want to able to do something like:
echo ${'count' . $arg1} #which should give me "test" but its returning bad-substitution error
i.e. can part of a variable can be set as variable in bash?
Okay, got it working via indirect variable reference introduced in Bash v2 as:
my_var="count$arg1"
echo ${!my_var}
Related
I'm having 2 Variables inside a variable group and their values are as shown below:
cr 2200
tr cd1200
I would like to assign these two variables to a third variable cd as shown below:
cd '2200;cd1200;abc.txt'
I'm trying to use below script but it is showing too many arguments.
Can someone please help in this.
I'm using below code:
cd = \'"$cr";"$tr";abc.txt\'
echo "$cd"
I need output as:
cd = '2200;cd1200;abc.txt'
Using Azure macro syntax ($(<varName>)):
$cd = '$(cr);$(tr);abc.txt'
Note:
PowerShell always needs the $ sigil when accessing a variable - even when assigning to it (unlike in POSIX-compatible shells such as Bash).
Azure's macro syntax - which textually expands references to Azure variables up front, before PowerShell sees the code, is not be confused with PowerShell's subexpression operator ($(...))
How can I retrieve a bash variable value if I have the variable name as string? var1="this is the real value"
a="var1"
Do something to get value of var1 just using variable a.
Context:
I have some AMI's (Amazon Machine Image) and I want to fire up a few instances of each AMI. As soon as they finish booting, I want to setup each instance according to its AMI type. I don't want to bake lots of scripts or secret keys inside any AMI so I prepared a generalized startup script and I put it on S3 with a publicly accessible link. In rc.local I put small piece of code which fetches the startup script and executes it. This is all I have in the AMIs. Then each AMI accesses a common configuration script which is applicable to all AMIs and special setup scripts for each. These scripts are private and require a signed URL to access them.
So now, when I fire an instance of an AMI (my_private_ami_1), I pass a signed URL for one more file presented on S3 which contains signed URL for all private scripts in terms of key/value pair.config_url="http://s3.amazo.../config?signature"
my_private_ami_1="http://s3.amazo.../ami_1?signature"
...
When the startup script runs, it downloads the above file and source's it. Then it checks for its AMI type and picks the correct setup script for itself.
ami\_type=GET AMI TYPE #ex: sets ami\_type to my\_private\_ami\_1
setup\_url=GET THE SETUP FILE URL BASED ON AMI\_TYPE # this is where this problem arises
So now I can have a generic code which can fire instances irrespective of their AMI types and instances can take care of themselves.
You can use ${!a}:
var1="this is the real value"
a="var1"
echo "${!a}" # outputs 'this is the real value'
This is an example of indirect parameter expansion:
The basic form of parameter expansion is ${parameter}. The value of
parameter is substituted.
If the first character of parameter is an exclamation point (!), it
introduces a level of variable indirection. Bash uses the value of the
variable formed from the rest of parameter as the name of the
variable; this variable is then expanded and that value is used in the
rest of the substitution, rather than the value of parameter itself.
X=foo
Y=X
eval "Z=\$$Y"
sets Z to foo.
For my fellow zsh users, the way to accomplish the same thing as the accepted answer is to use:
echo ${(P)a} # outputs 'this is the real value'
It is appropriately called Parameter name replacement
This forces the value of the parameter name to be interpreted as a
further parameter name, whose value will be used where appropriate.
Note that flags set with one of the typeset family of commands (in
particular case transformations) are not applied to the value of name
used in this fashion.
If used with a nested parameter or command substitution, the result of
that will be taken as a parameter name in the same way. For example,
if you have ‘foo=bar’ and ‘bar=baz’, the strings ${(P)foo},
${(P)${foo}}, and ${(P)$(echo bar)} will be expanded to ‘baz’.
Likewise, if the reference is itself nested, the expression with the
flag is treated as if it were directly replaced by the parameter name.
It is an error if this nested substitution produces an array with more
than one word. For example, if ‘name=assoc’ where the parameter assoc
is an associative array, then ‘${${(P)name}[elt]}’ refers to the
element of the associative subscripted ‘elt’.
Modified my search keywords and Got it :). eval a=\$$a Thanks for your time.
In bash 4.3+, you can use declare -n:
#!/usr/bin/env bash
var="this is the real value"
var_name="var"
declare -n var_ref=$var_name
echo "${var_ref}"
Had the same issue with arrays, here is how to do it if you're manipulating arrays too :
array_name="ARRAY_NAME"
ARRAY_NAME=("Val0" "Val1" "Val2")
ARRAY=$array_name[#]
echo "ARRAY=${ARRAY}"
ARRAY=("${!ARRAY}")
echo "ARRAY=${ARRAY[#]}"
echo "ARRAY[0]=${ARRAY[0]}"
echo "ARRAY[1]=${ARRAY[1]}"
echo "ARRAY[2]=${ARRAY[2]}"
This will output :
ARRAY=ARRAY_NAME[#]
ARRAY=Val0 Val1 Val2
ARRAY[0]=Val0
ARRAY[1]=Val1
ARRAY[2]=Val2
In bash 4.3, the '-v' test for set variables was introduced. At the same time, 'nameref' declaration was added. These two features together with the indirection operator (!) enable a simplified version of the previous example:
get_value()
{
declare -n var_name=$1
if [[ -v var_name ]]
then
echo "${var_name}"
else
echo "variable with name <${!var_name}> is not set"
fi
}
test=123
get_value test
123
test="\$(echo \"something nasty\")"
get_value test
$(echo "something nasty")
unset test
get_value test
variable with name <test> is not set
As this approach eliminates the need for 'eval', it is safer.
This code checked under bash 5.0.3(1).
modern shells already support arrays( and even associative arrays). So please do use them, and use less of eval.
var1="this is the real value"
array=("$var1")
# or array[0]="$var1"
then when you want to call it , echo ${array[0]}
Based on the answer: https://unix.stackexchange.com/a/111627
###############################################################################
# Summary: Returns the value of a variable given it's name as a string.
# Required Positional Argument:
# variable_name - The name of the variable to return the value of
# Returns: The value if variable exists; otherwise, empty string ("").
###############################################################################
get_value_of()
{
variable_name=$1
variable_value=""
if set | grep -q "^$variable_name="; then
eval variable_value="\$$variable_name"
fi
echo "$variable_value"
}
test=123
get_value_of test
# 123
test="\$(echo \"something nasty\")"
get_value_of test
# $(echo "something nasty")
VALUE=$(eval "echo \$$SOME_VAR_NAME")
Example
SSH_KEY_FILE_PATH_FOO="/tmp/key"
SSH_KEY_FILE_PATH_VAR_NAME_PREFIX="SSH_KEY_FILE_PATH"
SSH_KEY_FILE_PATH_VAR_NAME_SUFFIX="FOO"
SSH_KEY_FILE_PATH=$(eval "echo \$${SSH_KEY_FILE_PATH_VAR_NAME_PREFIX}_${SSH_KEY_FILE_PATH_VAR_NAME_SUFFIX}")
echo "$SSH_KEY_FILE_PATH"
/tmp/key
I am new to shell scripting. I have a very basic question about how we read variables defined in our terminal as input to a shell scirpt.
let us say i defined this variable in my terminal
a=22
if i do echo $a it gives 22 as output in my terminal.
I wanted to pass this variable as a parameter to a script named input.sh
#!/bin/bash
echo "Enter variable name:"
read Input
echo $Input
I ran the scirpt as ./input.sh It popped up the message as
Enter variable name:$a
But in the output i have $a as output not 22. I wanted 22 as output. Is there a way to do this?
You are mixing several things here. Where to start...
The usual way to pass values to a script is through positional parameters.
Suppose you have this script, called s1 :
#!/bin/bash
echo "First two args are: $1 $2"
If you execute it like this :
./s1 Arg1 Arg2
You will see the following output :
First two args are: Arg1 Arg2
If you want to pass a variable name to the script, and have that script output the value of this variable, then you must do two things. First, initialize the variable and export it so that it can be seen by the children processes (including the script you will call, which is a separate process unless called with source or .).
VAR="Some value"
export VAR
You can also do both in a single statement:
export VAR="Some value"
Then, adapt the script to perform an indirect access to the variable :
#!/bin/bash
echo "Value of variable named $1 : ${!1}"
Please note that while $1 means "the content of variable 1", ${!1} means "the content of the variable that is named $1". This is the indirect part.
Now, if you want to take it a step further, and allow the script to interactively read user input (not an argument) and use the value read as a variable name to expand, you would do something like this :
#!/bin/bash
echo "Please enter a variable name"
read VARNAME
echo "Value of variable named $VARNAME : ${!VARNAME}"
Using positional parameters makes the script easier to reuse in non-interactive scenarios, so reading user input should be limited to cases where it is necessary.
The above is to help understand the basics. If you move beyond toy scripts, you will need to understand the security implications of indirect access (especially if you allow user input). You will also need to validate positional parameters or user input are valid for your purpose (i.e. contains a valid variable name) so that you may have your script react appropriately. Well, you would probably need to check if positional parameters were even provided to begin with. All of this is doable in shell scripting, but is beyond the scope of a single question. In any case, checking input (and also errors) will be required if you intend to have robust scripts in situations where reliability is expected.
To use the variables defined in the terminal in your bash script -
Change last line of your script so that it looks like -
#!/bin/bash
echo "Enter variable name:"
read Input
echo ${!Input}
Run your script input.sh on the terminal as -
. input.sh
And finally, when you input the variable name, do not use $ sign. For e.g. -
6c4008a16b7c:~ z001lg8$ . input.sh
Enter variable name:
a
22
6c4008a16b7c:~ z001lg8$
Voila, you can now use the variables defined in the terminal in your script.
Explanation -
In your script, $Input is changed to ${!Input} so that the content of the user input(which is variable name) is echoed and not the variable name itself.
As explained by #Fred - $1 means "the content of variable 1", ${!1} means "the content of the variable that is named $1".
When script is run as . input.sh , it means that you are sourcing the script contents on the terminal. The . symbol is used for sourcing a command/script.
The $ sign is not required when the variable name is entered in terminal because ${!Input} format already takes into account that the value in Input variable is a variable name.
New to Unix not aware of the syntax structure so please excuse my syntax brevity.I am trying to copy a value of a variable and store that in another variable eg:
Two variables:
abc
bcd
Given:
abc=123
I want to copy the contents of abc i.e 123 in bcd. How to achieve this in Unix?
Earlier I was trying to copy the contents of abc in a .txt file which was working for me: see the code snippet below:
abc='123'
echo $abc >>/data/test/tt.txt
But know I want to copy them in another variable so I tried to do the following but was of no success.
abc='123'
test=`echo $abc>>bcd`
echo $test
Can you assist me in this?
Easy:
bcd="$abc"
For example:
abc="hello world"
The quotes there are necessary or else it will try to run a command named world with abc in its environment.
Actually, the quotes are not necessary (thanks to 1_CR for pointing this), but I like to add them for readability:
bcd=$abc
bcd="$abc"
They both do the same, exactly what you need.
Lastly, do not use single quotes, or else you will not get the value of the variable:
bcd='$abc'
Error! Now your bcd variable contains the literal value $abc.
How can I retrieve a bash variable value if I have the variable name as string? var1="this is the real value"
a="var1"
Do something to get value of var1 just using variable a.
Context:
I have some AMI's (Amazon Machine Image) and I want to fire up a few instances of each AMI. As soon as they finish booting, I want to setup each instance according to its AMI type. I don't want to bake lots of scripts or secret keys inside any AMI so I prepared a generalized startup script and I put it on S3 with a publicly accessible link. In rc.local I put small piece of code which fetches the startup script and executes it. This is all I have in the AMIs. Then each AMI accesses a common configuration script which is applicable to all AMIs and special setup scripts for each. These scripts are private and require a signed URL to access them.
So now, when I fire an instance of an AMI (my_private_ami_1), I pass a signed URL for one more file presented on S3 which contains signed URL for all private scripts in terms of key/value pair.config_url="http://s3.amazo.../config?signature"
my_private_ami_1="http://s3.amazo.../ami_1?signature"
...
When the startup script runs, it downloads the above file and source's it. Then it checks for its AMI type and picks the correct setup script for itself.
ami\_type=GET AMI TYPE #ex: sets ami\_type to my\_private\_ami\_1
setup\_url=GET THE SETUP FILE URL BASED ON AMI\_TYPE # this is where this problem arises
So now I can have a generic code which can fire instances irrespective of their AMI types and instances can take care of themselves.
You can use ${!a}:
var1="this is the real value"
a="var1"
echo "${!a}" # outputs 'this is the real value'
This is an example of indirect parameter expansion:
The basic form of parameter expansion is ${parameter}. The value of
parameter is substituted.
If the first character of parameter is an exclamation point (!), it
introduces a level of variable indirection. Bash uses the value of the
variable formed from the rest of parameter as the name of the
variable; this variable is then expanded and that value is used in the
rest of the substitution, rather than the value of parameter itself.
X=foo
Y=X
eval "Z=\$$Y"
sets Z to foo.
For my fellow zsh users, the way to accomplish the same thing as the accepted answer is to use:
echo ${(P)a} # outputs 'this is the real value'
It is appropriately called Parameter name replacement
This forces the value of the parameter name to be interpreted as a
further parameter name, whose value will be used where appropriate.
Note that flags set with one of the typeset family of commands (in
particular case transformations) are not applied to the value of name
used in this fashion.
If used with a nested parameter or command substitution, the result of
that will be taken as a parameter name in the same way. For example,
if you have ‘foo=bar’ and ‘bar=baz’, the strings ${(P)foo},
${(P)${foo}}, and ${(P)$(echo bar)} will be expanded to ‘baz’.
Likewise, if the reference is itself nested, the expression with the
flag is treated as if it were directly replaced by the parameter name.
It is an error if this nested substitution produces an array with more
than one word. For example, if ‘name=assoc’ where the parameter assoc
is an associative array, then ‘${${(P)name}[elt]}’ refers to the
element of the associative subscripted ‘elt’.
Modified my search keywords and Got it :). eval a=\$$a Thanks for your time.
In bash 4.3+, you can use declare -n:
#!/usr/bin/env bash
var="this is the real value"
var_name="var"
declare -n var_ref=$var_name
echo "${var_ref}"
Had the same issue with arrays, here is how to do it if you're manipulating arrays too :
array_name="ARRAY_NAME"
ARRAY_NAME=("Val0" "Val1" "Val2")
ARRAY=$array_name[#]
echo "ARRAY=${ARRAY}"
ARRAY=("${!ARRAY}")
echo "ARRAY=${ARRAY[#]}"
echo "ARRAY[0]=${ARRAY[0]}"
echo "ARRAY[1]=${ARRAY[1]}"
echo "ARRAY[2]=${ARRAY[2]}"
This will output :
ARRAY=ARRAY_NAME[#]
ARRAY=Val0 Val1 Val2
ARRAY[0]=Val0
ARRAY[1]=Val1
ARRAY[2]=Val2
In bash 4.3, the '-v' test for set variables was introduced. At the same time, 'nameref' declaration was added. These two features together with the indirection operator (!) enable a simplified version of the previous example:
get_value()
{
declare -n var_name=$1
if [[ -v var_name ]]
then
echo "${var_name}"
else
echo "variable with name <${!var_name}> is not set"
fi
}
test=123
get_value test
123
test="\$(echo \"something nasty\")"
get_value test
$(echo "something nasty")
unset test
get_value test
variable with name <test> is not set
As this approach eliminates the need for 'eval', it is safer.
This code checked under bash 5.0.3(1).
modern shells already support arrays( and even associative arrays). So please do use them, and use less of eval.
var1="this is the real value"
array=("$var1")
# or array[0]="$var1"
then when you want to call it , echo ${array[0]}
Based on the answer: https://unix.stackexchange.com/a/111627
###############################################################################
# Summary: Returns the value of a variable given it's name as a string.
# Required Positional Argument:
# variable_name - The name of the variable to return the value of
# Returns: The value if variable exists; otherwise, empty string ("").
###############################################################################
get_value_of()
{
variable_name=$1
variable_value=""
if set | grep -q "^$variable_name="; then
eval variable_value="\$$variable_name"
fi
echo "$variable_value"
}
test=123
get_value_of test
# 123
test="\$(echo \"something nasty\")"
get_value_of test
# $(echo "something nasty")
VALUE=$(eval "echo \$$SOME_VAR_NAME")
Example
SSH_KEY_FILE_PATH_FOO="/tmp/key"
SSH_KEY_FILE_PATH_VAR_NAME_PREFIX="SSH_KEY_FILE_PATH"
SSH_KEY_FILE_PATH_VAR_NAME_SUFFIX="FOO"
SSH_KEY_FILE_PATH=$(eval "echo \$${SSH_KEY_FILE_PATH_VAR_NAME_PREFIX}_${SSH_KEY_FILE_PATH_VAR_NAME_SUFFIX}")
echo "$SSH_KEY_FILE_PATH"
/tmp/key