Apache XML-RPC Exception Handling

Question

What is the easiest way to extract the original exception from an exception returned via Apache's implementation of XML-RPC?

What is the context of this question? Are you using XML-RPC directly? — ScArcher2, Sep 09 '08 at 21:44
I'm using Apache's implementation, namely http://ws.apache.org/xmlrpc/ — John with waffle, Sep 11 '08 at 14:04

John with waffle · Accepted Answer · 2008-12-28T02:39:50.057

4

It turns out that getting the cause exception from the Apache exception is the right one.

} catch (XmlRpcException rpce) {
    Throwable cause = rpce.getCause();
    if(cause != null) {
        if(cause instanceof ExceptionYouCanHandleException) {
            handler(cause);
        }
        else { throw(cause); }
    }
    else { throw(rpce); }
}

edited Dec 28 '08 at 02:39

answered Sep 11 '08 at 14:09

John with waffle

4,113
21
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score 1 · Answer 2 · answered Sep 09 '08 at 21:49

According to the XML-RPC Spec it returns the "fault" in the xml.

Is this the "Exception" you are referring to or are you refering to a Java Exception generated while making the XML-RPC call?

Fault example

HTTP/1.1 200 OK
Connection: close
Content-Length: 426
Content-Type: text/xml
Date: Fri, 17 Jul 1998 19:55:02 GMT
Server: UserLand Frontier/5.1.2-WinNT

<?xml version="1.0"?>
<methodResponse>
  <fault>
    <value>
      <struct>
      <member>
        <name>faultCode</name>
        <value><int>4</int></value>
      </member>
      <member>
        <name>faultString</name>
        <value>
          <string>Too many parameters.</string>
        </value>
      </member>
      </struct>
    </value>
  </fault>
</methodResponse>

Let's say basic authentication is used... how do you get the header to be HTTP/1.1 401 Unauthorized rather? — Jaco Van Niekerk, Jan 17 '12 at 12:24

Apache XML-RPC Exception Handling

2 Answers2