[SOLVED] shell command in variable

rohit_k · 06-12-2012, 03:36 AM

Code:

# a="xx : yy : cat /some/file | grep -i string"
# b=`echo $a | cut -d":" -f3` 

# echo "$b"
cat /some/file | grep -i string

I want to execute the command "cat /some/file | grep -i string"

Code:

# $b
cat: invalid option -- i
Try `cat --help' for more information.
#
#
# c=`$b`
cat: invalid option -- i
Try `cat --help' for more information.

how to get a command in a variable?

pan64 · 06-12-2012, 03:40 AM

what shell do you use?
You need only: grep -i string filename, you do not need cat at all

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rohit_k · 06-12-2012, 04:01 AM

Using /bin/bash (version: 3.2.25(1); OS is RHEl 5)
The problem is with | (pipe).
If i do

Code:

# grep -i 'someString' /some/file.xml
         <someString>[ipAddress]:port</someString>
#
# grep -i 'someString' /some/file.xml | sed 's/.* <someString>\(.*\)<\/someString>.*/\1/'
[ipAddress]:port
#
#a="grep -i 'someString' /some/file.xml | sed 's/.* <someString>\(.*\)<\/someString>.*/\1/'"
# $a
grep: |: No such file or directory
grep: sed: No such file or directory
grep: 's/.*: No such file or directory
grep: <someString>\(.*\)<\/someString>.*/\1/': No such file or directory

Pls help.

rknichols · 06-12-2012, 09:14 AM

Your problem is that bash performs variable expansion only after the pipelining has been parsed. The best solution is usually to build a shell function with the desired commands and pipelining instead of trying to store that in a variable:

Code:

function doit() {
    grep -i 'someString' /some/file.xml | sed 's/.* <someString>\(.*\)<\/someString>.*/\1/'
}
doit

David the H. · 06-12-2012, 10:18 AM

rknichols has provided the correct answer, but here's the documentation to back it up:

I'm trying to put a command in a variable, but the complex cases always fail!
http://mywiki.wooledge.org/BashFAQ/050

In a nutshell, variables are for storing data, not code. Use functions.

And BTW, $(..) is highly recommended over `..`.

rohit_k · 06-12-2012, 11:16 PM

Thnx rknichols and David.