Shell Programming and Scripting

Hi,

I am trying to set a variable to be used in later scripting, and am having some difficulty. I want to look in a file called scan.info and find the line that says "variable ok". Then I want to cut the number at the beginning of that line and assign that number as a variable so that later commands can refer back to ${variable}. I've tried both:

variable = $( grep "variable ok" scan.info | cut -c2-3 )
set variable = $( grep "variable ok" scan.info | cut -c2-3 )

I know that the part in parenthesis works when I type it into the terminal, but when I try to run it in a script with the set variable aspect added, it says:

Illegal variable name.

The part I'm most confused about is that the same command line works in a script used by a coworker, but not in my script.

Any ideas? Thanks!

Doing yours I got a 0 as the variable using the below file.

Here is how to do it:

Code:

me@hostname:~/tmp> cat tmp.sh
#!/bin/sh

var=$(grep "variable ok" scan.info | cut -d' ' -f 1)
echo $var
me@hostname:~/tmp> cat scan.info
10 variable ok
20 testing
30 test
me@hostname:~/tmp>

This way is assuming that the number you want is always at the beginning and there is always going to be a space right after it.

I tried that way, and it still didn't work. I'm using csh, could that be why?

My scan.info file is set up like:

1 test ok 512 512 3
2 testing ok 64 64 1
3 variable error 56 64 1
4 variable ok 56 64 1
etc

So I want to set alpha to the number 4. I tried:

alpha = $(grep "variable ok" scan.info | cut -d" " -f 1) but still got the Illegal variable name error.

Any idea?

I'm not very familiar with the c shell, but this seemed to work:

Code:

#!/bin/csh

set var=`grep "variable ok" scan.info | cut -d' ' -f 1`
echo $var

Thanks! I got it to work!