How to handle XML in Java with a simple access to the nodes?

Question

What best meets my requirements in creating and modifying XML?

Requirements:

checking whether a node exists through the entire path from the root
```
<root>
   <a>
      <b>
      </b>
   </a>
<root>
```
-> //root/a/b
creating XML nodes with easy access to them, like expressions in XPath.

What can be recommend?

You may write a simple helper class to accomplish your task around DOM. See this http://stackoverflow.com/a/8346867/851432 — Jomoos, Dec 01 '11 at 19:18

score 2 · Answer 1 · edited May 23 '17 at 12:20

2

You can use Java XPath API in javax.xml.xpath package. This is a tutorial

http://www.ibm.com/developerworks/library/x-javaxpathapi/index.html

And to create XML nodes using XPath check this out

edited May 23 '17 at 12:20

Community

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answered Nov 22 '11 at 15:11

Raihan

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score 0 · Answer 2 · edited Jul 13 '12 at 18:07

0

Perhaps this is overkill for you but XmlBeans from Apache is pretty cool and JAXB with Binders (which is perhaps a bit more modern) are both great options for modelling strictly defined XML (those with XSD Schema, DTD) in a Java friendly manner as simple beans. With a little extra coding, both can use XPath to reference elements within the XML data.

edited Jul 13 '12 at 18:07

Peter Mortensen

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answered Nov 22 '11 at 15:46

Yaneeve

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score 0 · Answer 3 · answered Nov 22 '11 at 20:59

0

If you can use Groovy, XmlSlurper plays nicely with Java, e.g. reading and updating. Groovy has its own expression language GPath, which has similar purpose as the more well-known XPath.

Curious for more? Visit Groovy XML processing.

answered Nov 22 '11 at 20:59

matsev

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How to handle XML in Java with a simple access to the nodes?

3 Answers3