Question

grep in a shell script

Asked by: aavula

I have to write a shell script that checks for the word SUCCESS in the log file . If It didnt find the word SUCCESS in the log file then an error message should go out.
My script:
#!/usr/bin/sh
Result=SUCCESS
File=`find /opt/dev/logs/ -name "integchk.log.*"`
if [  $Result !=`grep SUCCESS $File | awk '{print $3}' | sed s/://` ]  then
echo "the process is a failure"
fi
The problem is that once the grep didnt find the word SUCCESS ,there is nothing to compare.
Can an expret please suggest me a way.

Thanks

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Asked On
2007-05-24 at 13:19:59ID22593523
Tags

grep

,

shell

,

script

Topics

Shell Scripting

,

Unix Operating Systems

Participating Experts
2
Points
125
Comments
4

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Answers

 

by: ozoPosted on 2007-05-24 at 13:44:46ID: 19152635

if grep SUCCESS $File ; then echo $Result
else
echo "the process is a failure"
fi

 

by: TintinPosted on 2007-05-24 at 15:24:03ID: 19153302

I'll assume you don't want to see the match entries, so do

grep SUCCESS $File >/dev/null || echo "The process is a failure"

or if you have GNU grep, you can use the -q flag, ie:

grep -q SUCCESS $File  || echo "The process is a failure"

Why do you have the line

File=`find /opt/dev/logs/ -name "integchk.log.*"`

??

If this matches two of more files, the if test will fail.  Is there ever only one log file?  If there are multiple files, just change the grep to:

grep SUCCESS /opt/dev/logs/integchk.log.* >/dev/null || echo "The process is a failure"


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