CODEWITHSUNDEEP
javaxmlbeans
samson
samson

Reputation: 1623

comparing two XmlBeans Objects for equality

I've an XML Beans Interface called SynonymsRequest with:

public interface SynonymsRequest extends org.apache.xmlbeans.XmlObject {...}

I want to test two instances of SynonymsRequest for equality:

SynonymsRequest s1 = SynonymsRequest.Factory.newInstance();
s1.setQueryText("blub");
s1.setRequesterId(BigInteger.valueOf(1));       
SynonymsRequest s2 = SynonymsRequest.Factory.newInstance();
s2.setQueryText("guck");
s2.setRequesterId(BigInteger.valueOf(1));

I've tried the following:

  1. assertTrue(s1.equals(s2)); => assertion does not pass
  2. assertEquals(0, s1.compareTo(s2)); => throws ClassCastException
  3. assertEquals(0, s1.compareValue(s2)); => assertion does not pass (returns 2, not compareable)
  4. assertTrue(s1.valueEquals(s2)); => always returns true, no matter if the two instances are equal

So what is the proper way of doing this?

Upvotes: 1

Views: 1372

Answers (4)

Sbspider
Sbspider

Reputation: 399

Noticed this a while back - if the two objects have toString() methods generated when they were made, then you can to an .equals on the toString() methods of the objects. These can be compared with relativ ease, since they will check if the output xml is equivalent.

Upvotes: 0

Kevin Krouse
Kevin Krouse

Reputation: 610

XmlBeans doesn't support a deep comparison so you'll have to write your own. There was a thread on the dev mailing list a while ago about a schema-aware comparison, but I'm not sure anything became of it:

http://www.mail-archive.com/[email protected]/msg01960.html

Upvotes: 0

Francisco Paulo
Francisco Paulo

Reputation: 6322

If it doesn't impact the performance of your program, you could compare them like this:

assertTrue(s1.xmlText().equals(s2.xmlText()));

Otherwise, I guess you will have to write your own custom comparator.

Upvotes: 3

kan
kan

Reputation: 28961

As I understand, the comparison compares two simple values only. It cannot deduct your desired comparison algorithm.

Or I don't understand what exactly do you mean?

Upvotes: 1

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