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How to escape the 'pipe' command
Name: Peter Date: March 11, 2002 at 16:23:49 Pacific
Comment:
How do I evaluate a command that is stored within a variable ?
I have the following script: #!/bin/sh my_func_call=ls -l | more echo "Evaluating $my_func_call" $my_func_call
I tried placing single quotes (', not `) arount the statement. When my call is only ls -l, all it fine. The problem is that the pipe and the more is translated as a directory ?
I would very much appreciate by email. I will post the answer after figuring it out. Thanks
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