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Shell Programming and Scripting

String Compare

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# 1  
Old 06-15-2009
String Compare
Hi i have a bash script which calls a PHP script (comparison_cron.php).
The php script prints the string "no_data_to_retrieve" without quotes.
Why does the following bash script not leave the while loop?

Code:
#!/bin/bash

cronOutput=$(php /srv/www/vhosts/xyz/httpdocs/comparison_cron.php)
quitStatus="no_data_to_retrieve"

while [ "$cronOutput" != "$quitStatus" ];
do
   echo $cronOutput
   echo $quitStatus
   cronOutput=$(php /srv/www/vhosts/xyz/httpdocs/comparison_cron.php)
done
echo finished

It seems that although $cronOutput and $quitStatus have the same content ("no_data_to_retrieve") the comparison fails. Why??

Thanks in advance
# 2  
Old 06-15-2009
First try to debug yourself....
put set -x in 2nd line
and execute. If still in doubt share the output.
# 3  
Old 06-15-2009
Thanks a lot,

the problem was a \r at end of the php output.

Thanks for your help!
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