Unleashing the power of XPath and wildcards in AJAX

Can anyone explain why the variable objProperties, which contains an xpath with a wildcard, is coming up empty in this scenario?

function getXMLServerObject (httpType, cmd, isAsync) {
    var object = new Array();
    $.ajax({
      type: httpType,
      url: cmd,
      async: isAsync,
      dataType: "xml",
        success: function (xml) {
          var error = $(xml).find("Error");
          if (error == null) {
              object["__error"] = error.text();
              return object;
          }
          var objProperties = $(xml).find("Data\\Record\\*");

This results in an empty array for xml:

...

<Data>
    <Record>
        <MailFrequency>30</MailFrequency>
        <Tracking>-1</Tracking>
    </Record>
</Data>

However, searching for individual nodes returns the expected results.

javascriptajaxxmlxpathwildcard

Answer №1

jQuery does not support XPath selectors by default.

When you convert XML into a jQuery object, treat the tags like you would HTML tags and use the same jQuery selector methods.

If you have multiple <Record> elements, you can try this:

var $records = $(xml).find("Record"); /* create a collection of all Record tags */

$records.each(function(){
  var $children=$(this).children();
   /* or to access specific elements*/
   var mailFreq= $(this).find('MailFrequency').text();
})

Answer №2

When examining the code

var objProperties = $(xml).find("Data\\Record\\*");

It seems that the XPath expression "Data\\Record\\*" is being used, although it is unusual to have \ as a separator in XPath. Typically / is used: "Data/Record/*".

In addition, you can achieve the same result with jQuery selectors, which are an extension of CSS3 selectors. Remember that the root node is Data and $.find only searches among children:

var objProperties = $(xml).find("Record > *");

This will not create proper objects for the elements inside each record. To accomplish that, you should follow charlietfl's approach with some adjustments.

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