06-29-2017 01:48 AM
Hello,
i'm writing a Java Programm to add and remove membership of lines to Line Groups.
adding works well, but how can i find out the membership of a specific line?
i found the Element AXLSoap - LLineGroupMember , but it has no reference, so it seems that it exists, but it is not used by any AXL Request.
i tried the following without success:
AXLSoap - listLineGroup -> gives me no members
AXLSoap - getLineGroup -> gives me no members
please tell me that i don't have to use a sql search statement....
Solved! Go to Solution.
06-29-2017 08:47 AM
Hi r.rung,
Here is what I used to display the line members of a linegroup:
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns="http://www.cisco.com/AXL/API/11.0">
<soapenv:Header/>
<soapenv:Body>
<ns:getLineGroup sequence="?">
<name>LG_Name</name>
<returnedTags>
<members uuid="?">
<member uuid="?">
<directoryNumber uuid="?">
<pattern></pattern>
</directoryNumber>
</member>
</members>
</returnedTags>
</ns:getLineGroup>
</soapenv:Body>
</soapenv:Envelope>
Hope this helps
06-29-2017 08:47 AM
Hi r.rung,
Here is what I used to display the line members of a linegroup:
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:ns="http://www.cisco.com/AXL/API/11.0">
<soapenv:Header/>
<soapenv:Body>
<ns:getLineGroup sequence="?">
<name>LG_Name</name>
<returnedTags>
<members uuid="?">
<member uuid="?">
<directoryNumber uuid="?">
<pattern></pattern>
</directoryNumber>
</member>
</members>
</returnedTags>
</ns:getLineGroup>
</soapenv:Body>
</soapenv:Envelope>
Hope this helps
07-03-2017 04:02 AM
amazing Thanks.
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