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Say I declare variable var as var="ls | grep a". Then I would expect the command $var to work exactly as ls | grep a, but instead the | char is read as escaped \| so that it rather prints an error saying that no file named \|, grep or a is found.

My question is: how can I make Bash interpret | in var's value as not escaped?

I have to use it in a case statement such as

case $option in
  ${var1} ) echo "It works! 1" ;;
  ${var2} ) echo "It works! 2" ;;
  *  ) echo "It does not work"

(edit2): and make it return It works! 1 if $option is either str1 or str2, and It works! 2 if $option is either str3 or str4. That is because I am writing a function in which case patterns may change according to the user inputs.

Though I don't think it is relevant to know, I use lxterminal on Lubuntu 12.04 and $BASH_VERSION is 4.2.25(1)-release.

Any help is appreciated, thank you!

share|improve this question
It would be very helpful that you explain what your trying to achieve. It seems you're taking the wrong path when putting compound commands in a variable... there's probably a much more robust, safer, cleaner and easier way to achieve your goal. Please see this link about the XY problem. Besides, parsing the output of ls is a very bad thing to do! – gniourf_gniourf Oct 5 '13 at 17:50
I just want to have a case statement whose patterns depend on the values of some variables. In order to achieve this I thought it best to use variables as patterns, but the problem is that the vertical line is read as \|. Any solution is fine to me, I just wrote what I tried. – AndreasT Oct 5 '13 at 18:10
I really have no idea what you want to do and why. If you try case $var in *b*) echo "It works!";; it will work. – gniourf_gniourf Oct 5 '13 at 18:12
E.g. if var="a|b" I would want case $option in $var ) ... to produce the same result as case $option in a|b ) ..., if var="h|H|l|L I would want case $option in $var ) ... to produce the same result as case $option in h|H|l|L ) ... and so on. Briefly speaking, I'd like the "pattern )" to be variable. – AndreasT Oct 5 '13 at 18:18
For a menu, you can use the select bash builtin. help select for more info. This solution might be less clunky than what you're trying to do. – gniourf_gniourf Oct 6 '13 at 8:53
up vote 1 down vote accepted

If I understand your question, this may be what you wish:


first_option=$(echo $var | cut -d"|" -f1)
second_option=$(echo $var | cut -d"|" -f2)

is_working? () {
  case $1 in 
    $first_option ) echo "It works!" ;;
    $second_option ) echo "It also works!" ;;
    *  ) echo "It does not work"

is_working? $option

is_working? $option

is_working? $option
share|improve this answer
Thank you very much for your answer, but as @gniourf_gniourf commented I guess I wasn't very clear about what I meant. I would like the cases in the case statement to be variable as well as the option being tested. I slightly edited my question hoping it can clarify what I am looking for :) – AndreasT Oct 5 '13 at 19:27
@AndreasT Nothing more then simple, check my answer now. – Radu Rădeanu Oct 5 '13 at 19:42
Why reversing option and var? If var="aa|b" I would expect it "to work" if and only if $option is either aa or b but your first example would print It works! anyway. Besides, what if I had case ${option} in ${var1} )...;; ${var2} )...? – AndreasT Oct 5 '13 at 20:03
I mean, your code works only for a single char ${option} value – AndreasT Oct 5 '13 at 20:05
@AndreasT Ok, try it now – Radu Rădeanu Oct 5 '13 at 20:14

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