Luis,<br><br>forward() does not accept a variable for an argument.<br>instead use like:<br> $du = "sip:" + src_ip + ":5070";<br>I think it needs URI format without the user portion.<br>see: <a href="http://www.opensips.org/Resources/DocsCoreVar17#toc34">http://www.opensips.org/Resources/DocsCoreVar17#toc34</a><br>
<br>There are a bunch of older functions that do not accept variables as arguments. Some of them like this have read/write variables that can be used instead.<br>It would probably be good to farmiliarize yourself with all the vars on that page especially the read/write ones.<br>
<br>Dave<br><br><div class="gmail_quote">On Sun, Oct 2, 2011 at 7:17 PM, Luis Morales <span dir="ltr"><<a href="mailto:luisalfredo_ml31@hotmail.com">luisalfredo_ml31@hotmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
<div><div dir="ltr">
Greetings,<div><br></div><div>I'd like to use a variable with the function forward() to pass it the destination, but I can't find a way to do it. I've look in the documentation, but i don't know if I'm doing something wrong, or it just can't be done. </div>
<div><br></div><div>I'd like to something like this:</div><div><br></div><div>$var(dest) = src_ip + ":5070";</div><div>forward( $var(dest));</div><div><br></div><div><br>Thanks, <br><font color="#888888"><div>
Luis Morales.</div></font></div> </div></div>
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