jQuery XML parsing with namespaces

Question

I'm new to jQuery and would like to parse an xml document.

I'm able to parse regular XML with the default namespaces but with xml such as:

<xml xmlns:s="uuid:BDC6E3F0-6DA3-11d1-A2A3-00AA00C14882" xmlns:dt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882" xmlns:rs="urn:schemas-microsoft-com:rowset" xmlns:z="#RowsetSchema">
   <s:Schema id="RowsetSchema">
     <s:ElementType name="row" content="eltOnly" rs:CommandTimeout="30">
       <s:AttributeType name="ows_ID" rs:name="ID" rs:number="1">
        <s:datatype dt:type="i4" dt:maxLength="4" />
      </s:AttributeType>
       <s:AttributeType name="ows_DocIcon" rs:name="Type" rs:number="2">
        <s:datatype dt:type="string" dt:maxLength="512" />
      </s:AttributeType>
       <s:AttributeType name="ows_LinkTitle" rs:name="Title" rs:number="3">
        <s:datatype dt:type="string" dt:maxLength="512" />
      </s:AttributeType>
       <s:AttributeType name="ows_ServiceCategory" rs:name="Service Category" rs:number="4">
        <s:datatype dt:type="string" dt:maxLength="512" />
      </s:AttributeType>
    </s:ElementType>
  </s:Schema>
   <rs:data>
    <z:row ows_ID="2" ows_LinkTitle="Sample Data 1" />
    <z:row ows_ID="3" ows_LinkTitle="Sample Data 2" />
    <z:row ows_ID="4" ows_LinkTitle="Sample Data 3" />
  </rs:data>
</xml>

All I really want are the <z:row>'s.

So far, I've been doing:

$.get(xmlPath, {}, function(xml) {
    $("rs:data", xml).find("z:row").each(function(i) {
        alert("found zrow");
    });
}, "xml");

With really no luck. Any ideas? Thanks.

Answer 1

I have not seen any documentation on using JQuery to parse XML. JQuery typically uses the Browser dom to browse an HTML document, I don't believe it reads the html itself.

You should probably look at the built in XML handling in JavaScript itself.

http://www.webreference.com/programming/javascript/definitive2/

Answer 2

I got it.

Turns out that it requires \\ to escape the colon.

$.get(xmlPath, {}, function(xml) {
    $("rs\\:data", xml).find("z\\:row").each(function(i) {
        alert("found zrow");
    });
}, "xml");

As Rich pointed out:

The better solution does not require escaping and works on all "modern" browsers:

.find("[nodeName=z:row]")

Answer 3

Although the above answer seems to be correct, it does not work in webkit browsers (Safari, Chrome). A better solution I believe would be:

.find("[nodeName=z:row]")