Comparing XML equal files (except for whitespace) with xmlunit produces difference

Question

Using xmlunit (version 2.6.0) to compare two XML files produces differences when they should be considered equal.

XML 1:

<top><a>one</a><b>two</b></top>

XML 2:

<top>
  <a>one</a>
  <b>two</b>
</top>

Java code: Source xmlSource1 = Input.fromFile(xmlFile1).build(); Source xmlSource2 = Input.fromFile(xmlFile2).build();

DefaultNodeMatcher nodeMatcher = new DefaultNodeMatcher(ElementSelectors.byNameAndText);

Diff d = DiffBuilder.compare(xmlSource1)
            .withNodeMatcher(nodeMatcher)
            .withTest(xmlSource2).build();
    Iterable<Difference> diffList = d.getDifferences();
    Iterator<Difference> iterator = diffList.iterator();
    while(iterator.hasNext()) {
        Difference next = iterator.next();
        log.info("Difference: " + next);
    }

Produces this output:

Difference: Expected child 'top' but was 'null' - comparing <top...> at /top[1] to <NULL> (DIFFERENT)
Difference: Expected child 'null' but was 'top' - comparing <NULL> to <top...> at /top[1] (DIFFERENT)

Question: why are they considered different? How can this comparison be done by ignoring whitespace differences? Ideally I would like d.hasDifferences() to be false.

diginoise · Accepted Answer · 2018-07-19 16:57:40Z

3

Just ignore whitespaces (and comments) and perform check where similarities (see checkForSimilar()) are not included in the diff:

Diff d = DiffBuilder.compare(xmlSource1).withTest(xmlSource2)
 .checkForSimilar()
 .withNodeMatcher(nodeMatcher)
 .ignoreWhitespace()
 .ignoreComments()
 .build();

answered Jul 19, 2018 at 16:57

diginoise

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1 Comment

Barry S Over a year ago

Thanks for your quick response!

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