In my bash script I want to save the output of a cat command in a variable.I do the following:
it keeps returning error.why is that?Code:var1=`cat outtemp | sed '1d' ` echo $var1
the error is
var1=12866: Command not found.
var1: Undefined variable.
Last edited by quo; 05-30-2012 at 12:31 PM.
Not sure, it seems to work on my system:
Can you post your whole shell script? Also, post the contents of outtemp and any other files if possible.Code:$ cat outtemp this is a test $ var1=`cat outtemp | sed '1d' ` $ echo $var1 is a test
ok forget about the sed to make it more simple.Originally Posted by anduril462
In outtemp there's one string
John.Green
var1=`cat outtemp`
echo $var1
it keeps returning the same error
the simple cat command works but the problem occurs when I store it in the variable
ps.I have nothing else in my script,I just wanted to test this first
Last edited by quo; 05-30-2012 at 01:19 PM.
Still works for me on my system.
Describe your system for me. What OS and version? What shell and version are you using? How are you running this (i.e. from a shell script -- text file with #! at the top and execute permissions) or directly from the command line? Try it both ways. Do you get the same error both ways? What other info can you provide that would help me help you.
thank you so much for mentioning that because I just realized my stupid mistake,I forgot to put the !
I had instead #/bin/bash
Thanks a lot,can't believe it was just this
