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Old 05-12-2005, 08:36 AM   #1
jeffreybluml
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Registered: Mar 2004
Location: Minnesota
Distribution: Fedora Core 1, Mandrake 10
Posts: 405

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incrementing variable in an until loop


What's wrong with this?

Code:
x=1;
until $x=30;
do
grep blah blah blah blah blah...;
sed blah blah blah blah;
x=$x+1;
done;
Becuase this is what I get...

Code:
/bin/getphone: line 6: 1=30: command not found
/bin/getphone: line 6: 1+1=30: command not found
/bin/getphone: line 6: 1+1+1=30: command not found
/bin/getphone: line 6: 1+1+1+1=30: command not found
/bin/getphone: line 6: 1+1+1+1+1=30: command not found
/bin/getphone: line 6: 1+1+1+1+1+1=30: command not found
/bin/getphone: line 6: 1+1+1+1+1+1+1=30: command not found
I'm trying to fix a "script" from another thread, and I can't figure out how to run the commands within the loop 30 times and stop and exit...

Greatly appreciate any assistance...

Jeff
 
Old 05-12-2005, 08:41 AM   #2
trevelluk
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Registered: Nov 2003
Location: Bristol, UK
Distribution: Debian Lenny, Gentoo (at work)
Posts: 388

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You need to use let to do arithmetic in bash:

let "x = $x + 1"

That "until" syntax doesn't look right either. Try until [ $x -eq 30 ]; do
 
Old 05-12-2005, 08:45 AM   #3
jeffreybluml
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Registered: Mar 2004
Location: Minnesota
Distribution: Fedora Core 1, Mandrake 10
Posts: 405

Original Poster
Rep: Reputation: 30
Excellent!

Thank you!!

Now for the next problem....new thread...
 
  


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