Dear SIP: Thank your reply for my question. I modify my shell script and run the script alone. It can work

My shell script:

num=Bob@test.org if [ -z "$num" ] then echo "$*" else echo $num fi

When I call Alice@test1.org then request uri will change to Bob@test.org. But still did not work. Thank your response

Chungyu

That script a) has syntax errors, and b) doesn't make any sense.

Have you run the script alone? The line: if[! -z "$num" ] needs a space before the !

Also, logically, I don't get what you're trying to do. Currently, your script checks to see if $num has a value. If it DOES, then it echoes back the script arguments. If not, it echoes back sip:$num.... which doesn't make a bit of sense. You set $num in the beginning, so it ALWAYS has a value... so it's always going to echo back the script arguments. And if, for some strange reason, it DIDN'T have a value, why would you want to echo back a non-value?

I don't get it.

chungyu wrote:

...

Dear all: I want to change the content of request uri. I use this function ,exec_dset(), to change it. This is my configuration: if (!lookup("location")) { exec_dset("/usr/local/etc/openser/shell/sh1"); #sl_send_reply("404", "Not Found"); exit; }; shell scrip: sh1 #!/bin/sh num=200@xx.xx.xx.xx mailto:num=200@xx.xx.xx.xx if [! -z "$num" ] then echo "$*" else echo "sip:$num" fi

but it doesn't work. error message: ERROR:exec_str: no uri from /usr/local/etc/openser/shell/sh1 XX@xx.xx.xx mailto:XX@xx.xx.xx..... Can somebody tell me what's wrong with my shell script. Thank you for your response chungyu 2007/06/4

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