Parsing XML file using Java [closed]

Question

I need to parse am XML file in Java and check some support based on the condition. I tried using the DOM for this.

  <Config>
      <Configuration>
        <Descriptor>
          <e id="0">
            <family>Rest</family>
            <supportedList>
              <_length>5</_length>
              <_type>TypeX[]</_type>
              <e id="0">value-1</e>
              <e id="1">value-2</e>
              <e id="2">value-3</e>
              <e id="3">value-4</e>
              <e id="4">value-5</e>
            </supportedList>
          </e>
          <e id="1">
            <family>Rest1</family>
            <supportedList>
              <_length>5</_length>
              <_type>TypeX[]</_type>
              <e id="0">value1-1</e>
              <e id="1">value1-2</e>
              <e id="2">value1-3</e>
              <e id="3">value1-4</e>
              <e id="4">value1-5</e>
            </supportedList>
          </e>
          </Descriptor>
      </Configuration>
    </Config>

In the above XML file,I need to go through the "e" block and then Descriptor block, and then search for Family - Rest1 and then iterate through supportedList. Check if supportedList contains user provided value if it exists then return true.

I tried using a DOM parser but I am not able to parse the file properly. The actual XML file that I am working on, is of 20K lines and has too much data. Any easy solution on this.

the actual file has quite a sensitive data and the same thing I have used it in my code as well so could not share the code. — anirudha sonar
– anirudha sonar, Commented Apr 16, 2013 at 10:51

Joop Eggen · Accepted Answer · 2013-04-16 11:02:51Z

0

    XPathFactory factory = XPathFactory.newInstance();
    XPath xpath = factory.newXPath();
    xPathExpression = xpath.compile("//family[text()='Rest1']/e");

    NodeList list = (NodeList) xPathExpression
        .evaluate(xml, XPathConstants.NODESET);

And on XPath syntax, it suffices to search for "XPath cheat sheet" or so.

answered Apr 16, 2013 at 11:02

Joop Eggen

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