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(a)test.org
if [ -z "$num" ]
then
echo "$*"
else
echo $num
fi
When I call Alice(a)test1.org then request uri will change to Bob(a)test.org.
But still did not work.
Thank your response
Chungyu
That script a) has syntax errors, and b) doesn't
make any sense.
Have you run the script alone? The line:
if[! -z "$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 $num in the beginning, so it ALWAYS
has a value... so it's always going to echo back the script arguments.
And if, for some strange reason, it DIDN'T have a value, why would you
want to echo back a non-value?
I don't get it.
chungyu wrote:
Dear all:
I want to change the content of request uri. I use this function
,exec_dset(), to change it.
This is my configuration:
if (!lookup("location")) {
exec_dset("/usr/local/etc/openser/shell/sh1");
#sl_send_reply("404", "Not Found");
exit;
};
shell scrip: sh1
#!/bin/sh
num=200(a)xx.xx.xx.xx <mailto:num=200@xx.xx.xx.xx>
if [! -z "$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(a)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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