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Old 04-29-2009, 05:30 AM   #1
shreyas_1
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Problem with exit command


Hi,

I am executing script written by other user.
I am executing script A.ksh, and A.ksh calls B.ksh.

But It is giving error:
/home/user/B.ksh[132]: exit: -1: unknown option

When I checked B.ksh
It has following lines of code
if [ $success = 0 ]
then
...
some action
...
else
...
some action
...
exit -1
fi


I am new to linux, Can you please help me?

Thank you
 
Old 04-29-2009, 05:55 AM   #2
ilikejam
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Hi.

Unix exit status codes have to be in the range 0-255, so -1 is out of range. I'd probably just change it to '1'.

Dave
 
Old 04-29-2009, 06:01 AM   #3
JulianTosh
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I agree with Dave. Typically, '-1' is meant to set all bits to 1 to flag a general error condition. Exiting with '255' would be probably be a better direct translation, but unless some other process is checking for a specific return value, anything other than zero would suffice.
 
Old 04-29-2009, 06:42 AM   #4
linuxlover.chaitanya
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Something that can help you further.

http://tldp.org/LDP/abs/html/exit-status.html
 
Old 04-29-2009, 07:45 AM   #5
colucix
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I agree. However it should not give an error, but it should translate it to 255. The problem is that it interprets the value -1 as an option. Eventually you can try
Code:
exit -- -1
but take in mind that the shell sees the exit code as 255.
 
Old 04-29-2009, 08:08 AM   #6
barunparichha
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[$success = 0]
The code always goes to else part and runs " exit -1".
 
Old 04-29-2009, 08:29 AM   #7
chrism01
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The above advice is correct.
In a addition, the code you've shown does not set success to anything, so the condition will fail, as it does.
Strictly speaking (as you seem to be doing a numerical comparison), the syntax is
Code:
if [[ $success -eq 0 ]]
then
...
as per http://www.tldp.org/LDP/abs/html/tes...ml#DBLBRACKETS
 
  


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