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POWhy does set -e cause my script to exit when it encounters the following?
 

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  1. POWhy does set -e cause my script to exit when it encounters the following?
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    <p>I have a bash script that checks some log files created by a cron job that have time stamps in the filename (down to the second). It uses the following code:</p> <pre><code>CRON_LOG=$(ls -1 $LOGS_DIR/fetch_cron_{true,false}_$CRON_DATE*.log 2&gt; /dev/null | sed 's/^[^0-9][^0-9]*\([0-9][0-9]*\).*/\1 &amp;/' | sort -n | cut -d ' ' -f2- | tail -1 ) if [ -f "$CRON_LOG" ]; then printf "Checking $CRON_LOG for errors\n" else printf "\n${txtred}Error: cron log for $CRON_NOW does not exist.${txtrst}\n" printf "Either the specified date is too old for the log to still be around or there is a problem.\n" exit 1 fi CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code") if [ -z "$CRIT_ERRS" ]; then printf "%74s[${txtgrn}PASS${txtrst}]\n" else printf "%74s[${txtred}FAIL${txtrst}]\n" printf "Critical errors detected! Outputting to console...\n" echo $CRIT_ERRS fi </code></pre> <p>So this bit of code works fine, but I'm trying to clean up my scripts now and implement set -e at the top of all of them. When i do it to this script, it exits with error code 1. Note that I have errors form the first statement dumping to /dev/null. This is because some days the file has the word "true" and other days "false" in it. Anyway, i don't think this is my problem because the script outputs "Checking xxxxx.log for errors." before exiting when I add set -e to the top.</p> <p>Note: the $CRON_DATE variable is derived form user input. I can run the exact same statement from command line "$./checkcron.sh 01/06/2010" and it works fine without the set -e statement at the top of the script. </p> <p>UPDATE: I added "set -x" to my script and narrowed the problem down. The last bit of output is:</p> <pre><code>Checking /map/etl/tektronix/logs/fetch_cron_false_010710054501.log for errors ++ cat /map/etl/tektronix/logs/fetch_cron_false_010710054501.log ++ grep ERROR ++ grep -v 'Duplicate tracking code' + CRIT_ERRS= [1]+ Exit 1 ./checkLoad.sh... </code></pre> <p>So it looks like the problem is occurring on this line:</p> <pre><code>CRIT_ERRS=$(cat $CRON_LOG | grep "ERROR" | grep -v "Duplicate tracking code") </code></pre> <p>Any help is appreciated. :)</p> <p>Thanks, Ryan</p>
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    1. USSDGuero
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      1. POWhy does set -e cause my script to exit when it encounters the following?
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      1. VTUpMod
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      1. POWhy does set -e cause my script to exit when it encounters the following?
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      1. VTUpMod
    1. COParsing the output of ls in a script is generally considered a bad idea, because it does more than you want. In this case, it looks like you want ls to just print out the names of the files matched by the glob, one filename per line, and do nothing else. This can be accomplished with printf: printf '%s\n' $LOGS_DIR/fetch_cron_{true,false}_$CRON_DATE*.log This avoids any unpredicted behavior by ls, and also makes the script more portable.
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      1. POWhy does set -e cause my script to exit when it encounters the following?
      1. USEvan Krall
    2. COSee my edited answer.
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      1. POWhy does set -e cause my script to exit when it encounters the following?
      1. USDennis Williamson
    3. COAlso, instead of `cat filename|grep string` you should just do `grep string filename`
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      1. POWhy does set -e cause my script to exit when it encounters the following?
      1. USDennis Williamson
 

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