Assigning dynamic bash variable names using a for loop seq - linux

So I'm trying to do something, not sure if it's possible. I have the following code:
for i in {0..5}; do
if [[ -f ./user$i ]]; then
group$i=$(grep -w "group" ./user0|awk '{print $2}'|perl -lape 's/\s+//sg')
What I want to do is assign a unique variable for each instance of the {0..5} so group1 group2 group3 group4 for each variable name. Then I would change ./user0 to ./user$i and create a dynamic list of variables based on my sequence.
Is this possible? I get the following error when trying to execute this and I'm unsure of what I have actually done that bash doesn't like.
test.sh: line 16: group0=j: command not found

Kurt Stutsman provides the right pointer in a comment on the question: use Bash arrays to solve your problem.
Here's a simplified example:
groups=() # declare an empty array; same as: declare -a groups
for i in {0..5}; do
groups[i]="group $i" # dynamically create element with index $i
done
# Print the resulting array's elements.
printf '%s\n' "${groups[#]}"
See the bottom of this answer for other ways to enumerate the elements of array ${groups[#]}.
bash arrays can be dynamically expanded (and can even be sparse - element indices need not be contiguous)
Hence, simply assigning to element $i works, without prior sizing of the array.
Note how $i need not be prefixed with $ in the array subscript, because array subscripts are evaluated in an arithmetic context (the same context in which $(( ... )) expressions are evaluated).
As for what you did wrong:
group$i=...
is not recognized as a variable assignment by Bash, because - taken literally - group$i is not a valid identifier (variable name).
Because it isn't, Bash continues to parse until the next shell metacharacter is found, and then interprets the resulting word as a command to execute, which in your case resulted in error message group0=j: command not found.
If, for some reason, you don't want to use arrays to avoid this problem entirely, you can work around the problem:
By involving a variable-declaring builtin [command] such as declare, local, or export, you force Bash to perform expansions first, which expands group$i to a valid variable name before passing it to the builtin.
user2683246's answer demonstrates the next best approach by using declare (or, if local variables inside a function are desired, local) to create the variables.
Soren's answer uses export, but that is only advisable if you want to create environment variables visible to child processes rather than mere shell variables.
Caveat: With this technique, be sure to double-quote the RHS in order to capture the full value; to illustrate:
i=0; declare v$i=$(echo 'hi, there'); echo "$v0" # !! WRONG -> 'hi,': only UP TO 1ST SPACE
i=0; declare v$i="$(echo 'hi, there')"; echo "$v0" # OK -> 'hi, there'
Other ways to enumerate the groups array created above:
# Enumerate array elements directly.
for element in "${groups[#]}"; do
echo "$element"
done
# Enumerate array elements by index.
for (( i = 0; i < ${#groups[#]}; i++ )); do
echo "#$i: ${groups[i]}"
done

Use declare group$i=... instead of just group$i=...

Try to use the export or declare function like this
for i in {0..5}; do
if [[ -f ./user$i ]]; then
export group$i=$(grep -w "group" ......
with declare
for i in {0..5}; do
if [[ -f ./user$i ]]; then
declare group$i=$(grep -w "group" ......
where export makes the value available to sub-processes, and declare just available within the same script.

Related

Difference between "${param[0]}" and ${1} in bash

I'm looking at some old scripts and I found some parameter assignment that I have not seen before. A while loop reads from a text file and passes the values to a function. The items in the text file look like this:
user_one:abcdef:secretfolder
the first stage of the function then looks like this:
IFS=':' read -a param <<< $#
user="${param[0]}"
pass="${param[1]}"
user_folders="${param[2]}"
I have not seen this sort of assignment before and was wondering if this is just an alternative way of handling it. Is the above the same as this?
IFS=':' read -a param <<< $#
user="${1}"
pass="${2}"
user_folders="${3}"
(change in values to 1-3 due to ${0} being the name of the file itself). This script is 5 years old; This original sort of assignment just seems a longer way to to it, unless I've missed something
I'm still learning shell scripting but as I understand, setting IFS=':' will split the fields on : rather than whitespace and so in the examples, the value of "${param[0]}" and ${1} passed to the function would be user_one
Can someone please explain if there is a reason why "${param[0]}" should be used instead of ${1}?
The command:
IFS=':' read -a param <<< $#
reads the :-separated fields from the command arguments ($#) into the array variable named param. Bash arrays work just like lists in other languages, and you index them with brackets. ${param[0]} is the first field, ${param[1]} then next, and so on. Arrays like this can contain anything, and it's just because of the $# in the read command that this param array happens to contain the arguments. It could just as easily contain foo, bar, and baz if it were created like:
param=(foo bar baz)
The ${1}, ${2} etc. syntax always refers to the script arguments though.

Bash - Get the VALUE of 'nested' variable into another variable [Edit: Indirect Variable Expansion] [duplicate]

This question already has answers here:
Bash - variable variables [duplicate]
(4 answers)
Lookup shell variables by name, indirectly [duplicate]
(5 answers)
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Is it possible to build variable names from other variables in bash? [duplicate]
(7 answers)
Closed 5 years ago.
I'm trying to get get the VALUE of a 'nested' variable into another variable and/or use the value directly as shown below
Below is an example scenario which exactly explains where I'm stuck
$ USER1_DIR=./user1/stuff
$ USER2_DIR=./user2/stuff
$ USER3_DIR=./user3/stuff
#User will be taken as input, for now assuming user is USER1
$ USER="USER1"
$ DIR=${USER}_DIR
$ echo $DIR
>> USER1_DIR
$ DIR=${${USER}_DIR}
>> -bash: ${${USER}_DIR}: bad substitution
Challenge 1:
Get DIR value to ./user1/stuff when the input is USER1
or
Get ./user1/stuff as output when the input is USER1
After I'm able to get through Challenge 1, I've to add some content to a file in the user directory like below
Desired output is as below
$ echo "Some stuff of user1" >> $DIR/${DOC}$NO
# Lets say DOC="DOC1" and NO="-346"
# So the content has to be added to ./user1/stuff/DOC1-346
# Assume that all Directories exists
FYI, The above code will be a part of a function in a bash script and it will be executed only on a Linux server.
Note : I don't know what to call variable DIR hence used the term 'nested' variable. It would be great to know what is it called, greatly appreciate any insight. :)
You can use eval, variable indirection ${!...}, or reference variables declare -n.
In the following, I will use lowercase variable names, since uppercase variable names are special by convention. Especially overwriting $USER is bad, because that variable normally contains your user name (without explicitly setting it). For the following code fragments assume the following variables:
user1_dir=./user1/stuff
user=user1
Eval
eval "echo \${${user}_dir}"
# prints `./user1/stuff`
Eval is a bash built-in that executes its arguments as if they were entered in bash itself. Here, eval is called with the argument echo "${user1_dir}".
Using eval is considered bad practice, see this question.
Variable Indirection
When storing the name of variable var1 inside another variable var2, you can use the indirection ${!var2} to get the value of var1.
userdir="${user}_dir"
echo "${!userdir}"
# prints `./user1/stuff`
Reference Variables
Instead of using indirection every time, you also can declare a reference variable in bash:
declare -n myref="${user}_dir"
The reference can be used similar to variable indirection, but without having to write the !.
echo "$myref"
# prints `./user1/stuff`
Alternatives
Your script may become easier when using (associative) arrays. Arrays are variables that store multiple values. Single values can be accessed by using an index. Normal arrays use natural numbers as indices. Associative arrays use arbitrary strings as indices.
(Normal) Arrays
# Create an array with three entries
myarray=(./user1/stuff ./user2/stuff ./user3/stuff)
# Get the first entry
echo "${myarray[0]}"
# Get the *n*-th entry
n=2
echo "${myarray[$n]}"
Associative Arrays
Declare an associative array with three entries
# Create an associative array with three entries
declare -A myarray
myarray[user1]=./user1/stuff
myarray[user2]=./user2/stuff
myarray[user3]=./user3/stuff
# Get a fixed entry
echo "${myarray[user1]}"
# Get a variable entry
user=user1
echo "${myarray[$user]}"

How can I know if a string contains only one or several words in Bash? [duplicate]

This question already has answers here:
A confusion about ${array[*]} versus ${array[#]} in the context of a bash completion
(2 answers)
Closed 6 years ago.
When I get the content of an array in a string, I have the 2 solutions bellow :
$ a=('one' 'two')
$ str1="${a[*]}" && str2="${a[#]}"
After, of course, I can reuse my string on the code
but how can I know if my variable has only one or several words?
In both cases, the contents of the array are concatenated to a single string and assigned to the variable. The only difference is what is used to join the elements. With ${a[*]}, the first character of IFS is used. With ${a[#]}, a single space is always used.
$ a=(one two)
$ IFS="-"
$ str1="${a[*]}"
$ str2="${a[#]}"
$ echo "$str1"
one-two
$ echo "$str2"
one two
When expanding $str1 or $str2 without quoting, the number of resulting words is entirely dependent on the current value of IFS, regardless of how the variables were originally defined. "$str1" and "$str2" each expand, of course, to a single word.
To add to #chepner's great answer: the difference between ${arr[*]} and ${arr[#]} is very similar to the difference between $* and $#. You may want to refer to this post which talks about $* and $#:
What's the difference between $# and $* in UNIX?
As a rule of thumb, it is always better to use "$#" and "${arr[#]}" than their unquoted or * counterparts.
"${a[*]}" expands to one string for all entries together and "${a[#]}" expands to one string per entry.
Assume we had a program printParameters, which prints for each parameter ($1, $2, and so on) the string my ... parameter is ....
>_ a=('one' 'two')
>_ printParameters "${a[*]}"
my 1. parameter is one two
>_ printParameters "${a[#]}"
my 1. parameter is one
my 2. parameter is two
If you would expand the array manually, you would write
${a[*]} as "one two" and
${a[#]} as "one" "two".
There also differences regarding IFS and so on (see other answers). Usually # is the better option, but * is way faster – use the latter one in cases where you have to deal with large arrays and don't need separate arguments.
By the way: The script printParameters can be written as
#! /bin/bash
argIndex=0
for argValue in "$#"; do
echo "my $((++i)). argument is $argValue"
done
It's very useful for learning more about expansion by try and error.

bash getopts - difference between ${OPTARG} and $OPTARG [duplicate]

In shell scripts, when do we use {} when expanding variables?
For example, I have seen the following:
var=10 # Declare variable
echo "${var}" # One use of the variable
echo "$var" # Another use of the variable
Is there a significant difference, or is it just style? Is one preferred over the other?
In this particular example, it makes no difference. However, the {} in ${} are useful if you want to expand the variable foo in the string
"${foo}bar"
since "$foobar" would instead expand the variable identified by foobar.
Curly braces are also unconditionally required when:
expanding array elements, as in ${array[42]}
using parameter expansion operations, as in ${filename%.*} (remove extension)
expanding positional parameters beyond 9: "$8 $9 ${10} ${11}"
Doing this everywhere, instead of just in potentially ambiguous cases, can be considered good programming practice. This is both for consistency and to avoid surprises like $foo_$bar.jpg, where it's not visually obvious that the underscore becomes part of the variable name.
Variables are declared and assigned without $ and without {}. You have to use
var=10
to assign. In order to read from the variable (in other words, 'expand' the variable), you must use $.
$var # use the variable
${var} # same as above
${var}bar # expand var, and append "bar" too
$varbar # same as ${varbar}, i.e expand a variable called varbar, if it exists.
This has confused me sometimes - in other languages we refer to the variable in the same way, regardless of whether it's on the left or right of an assignment. But shell-scripting is different, $var=10 doesn't do what you might think it does!
You use {} for grouping. The braces are required to dereference array elements. Example:
dir=(*) # store the contents of the directory into an array
echo "${dir[0]}" # get the first entry.
echo "$dir[0]" # incorrect
You are also able to do some text manipulation inside the braces:
STRING="./folder/subfolder/file.txt"
echo ${STRING} ${STRING%/*/*}
Result:
./folder/subfolder/file.txt ./folder
or
STRING="This is a string"
echo ${STRING// /_}
Result:
This_is_a_string
You are right in "regular variables" are not needed... But it is more helpful for the debugging and to read a script.
Curly braces are always needed for accessing array elements and carrying out brace expansion.
It's good to be not over-cautious and use {} for shell variable expansion even when there is no scope for ambiguity.
For example:
dir=log
prog=foo
path=/var/${dir}/${prog} # excessive use of {}, not needed since / can't be a part of a shell variable name
logfile=${path}/${prog}.log # same as above, . can't be a part of a shell variable name
path_copy=${path} # {} is totally unnecessary
archive=${logfile}_arch # {} is needed since _ can be a part of shell variable name
So, it is better to write the three lines as:
path=/var/$dir/$prog
logfile=$path/$prog.log
path_copy=$path
which is definitely more readable.
Since a variable name can't start with a digit, shell doesn't need {} around numbered variables (like $1, $2 etc.) unless such expansion is followed by a digit. That's too subtle and it does make to explicitly use {} in such contexts:
set app # set $1 to app
fruit=$1le # sets fruit to apple, but confusing
fruit=${1}le # sets fruit to apple, makes the intention clear
See:
Allowed characters in Linux environment variable names
The end of the variable name is usually signified by a space or newline. But what if we don't want a space or newline after printing the variable value? The curly braces tell the shell interpreter where the end of the variable name is.
Classic Example 1) - shell variable without trailing whitespace
TIME=10
# WRONG: no such variable called 'TIMEsecs'
echo "Time taken = $TIMEsecs"
# What we want is $TIME followed by "secs" with no whitespace between the two.
echo "Time taken = ${TIME}secs"
Example 2) Java classpath with versioned jars
# WRONG - no such variable LATESTVERSION_src
CLASSPATH=hibernate-$LATESTVERSION_src.zip:hibernate_$LATEST_VERSION.jar
# RIGHT
CLASSPATH=hibernate-${LATESTVERSION}_src.zip:hibernate_$LATEST_VERSION.jar
(Fred's answer already states this but his example is a bit too abstract)
Following SierraX and Peter's suggestion about text manipulation, curly brackets {} are used to pass a variable to a command, for instance:
Let's say you have a sposi.txt file containing the first line of a well-known Italian novel:
> sposi="somewhere/myfolder/sposi.txt"
> cat $sposi
Ouput: quel ramo del lago di como che volge a mezzogiorno
Now create two variables:
# Search the 2nd word found in the file that "sposi" variable points to
> word=$(cat $sposi | cut -d " " -f 2)
# This variable will replace the word
> new_word="filone"
Now substitute the word variable content with the one of new_word, inside sposi.txt file
> sed -i "s/${word}/${new_word}/g" $sposi
> cat $sposi
Ouput: quel filone del lago di como che volge a mezzogiorno
The word "ramo" has been replaced.

Default values for the arguments to a Unix shell script?

Normally when a parameter is passed to a shell script, the value goes into ${1} for the first parameter, ${2} for the second, etc.
How can I set the default values for these parameters, so that if no parameter is passed to the script, we can use a default value for ${1}?
You can't, but you can assign to a local variable like this: ${parameter:-word} or use the same construct in the place you need $1. this menas use word if _paramater is null or unset
Note, this works in bash, check your shell for the syntax of default values
You could consider:
set -- "${1:-'default for 1'}" "${2:-'default 2'}" "${3:-'default 3'}"
The rest of the script can use $1, $2, $3 without further checking.
Note: this does not work well if you can have an indeterminate list of files at the end of your arguments; it works well when you can have only zero to three arguments.
#!/bin/sh
MY_PARAM=${1:-default}
echo $MY_PARAM
Perhaps I don't understand your question well, yet I would feel inclined to solve the problem in a less sophisticated manner:
! [[ ${1} ]] && declare $1="DEFAULT"
Hope that helps.

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