The script, exactly as you give it here fails with a syntax error in the test command because you initialize "change" to 0 but then use $changed. This, in /bin/sh will always cause a syntax error because $changed has no value and the "-eq" is a binary operator.
I am unable to test your script on a Linux machine in true "sh" because all of my systems have /bin/sh linked to /bin/bash. I was able to try it on a TRU-64 (osf5.1) UNIX machine with a real "bin/sh".
Results: Original script fails in both bash and true sh
After taking care of the apparent typo by changing "change=0" to "changed=0" the script works in both environments. The final echo emits 1.
If your typo was only in the mail and not in the real script, the I have no idea why it wouldn't work in /bin/sh on Linux.
Dave Basener
Brian McGrew wrote:
I'm asking this question here becuase what I have works on Solaris but not Redhat 7.3
I have a simple script:
#!/bin/sh
change=0
while [ $changed -eq 0 ]; do changed=1 echo $changed done
echo $changed exit
When I come to the last echo $changed, it's back to 0. Any idea why that is? This works find on Solaris???
-brian
Brian D. McGrew { brian@xxxxxxxxxxxxxxxxxxx || brian@xxxxxxxxxxxxx }
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