I have to xml files say abc.xml & 123.xml which are almost similar, i mean has the same content, but the second one i.e, 123.xml has more content than the earlier one. I want to read both the files using Java, and compare whether the content present in abc.xml for each tag is same as that in 123.xml, something like object comparison. Please suggest me how to read the xml file using java and start comparing.
Thanks.
if you just want to compare then use this:
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true);
dbf.setCoalescing(true);
dbf.setIgnoringElementContentWhitespace(true);
dbf.setIgnoringComments(true);
DocumentBuilder db = dbf.newDocumentBuilder();
Document doc1 = db.parse(new File("file1.xml"));
doc1.normalizeDocument();
Document doc2 = db.parse(new File("file2.xml"));
doc2.normalizeDocument();
Assert.assertTrue(doc1.isEqualNode(doc2));
else see this http://xmlunit.sourceforge.net/
I would go for the XMLUnit. The features it provides :
Good Luck!
I would use JAXB to generate Java objects from the XML files and then compare the Java files. They would make the handling much easier.
In general, if you know that you have two files with identical structure but slightly different and unordered content you are going to have to "read" the files to compare the contents.
If you have the XML Schema for your XML files then you could use JAXB to create a set of classes that will represent the specific DOM that is defined by your XML schema. The benefit of this approach is that you will not have to parse the XML file through generic functions for elements and attributes but rather through the actual fields that make sense to your problem.
Of course, to be able to detect the presence of the same entry across both files you are going to have to "match" them through some common field (for example, some ID).
To help you with the duplicates discovery process you could use some relevant data structure from Java's collections, like the Set (or one of its derivatives)
I hope this helps.
The right approach depends on two factors:
(a) how much control do you want over how the comparison is done? For example, do you need to control whether whitespace is significant, whether comments should be ignored, whether namespace prefixes should be ignored, whether redundant namespace declarations should be ignored, whether the XML declaration should be ignored?
(b) what answer do you want? (i) a boolean: same/different, (ii) a list of differences suitable for a human to process, (iii) a list of differences suitable for an application to process.
The two techniques I use are: (a) convert both files to Canonical XML and then compare strings. This gives very little control and only gives a boolean result. (b) compare the two trees using the XPath 2.0 deep-equal() function or the extended Saxon version saxon:deep-equal(). The Saxon version gives more control over how the comparison is done, and a more detailed report of the differences found (for human reading, not for application use).
If you want to write Java code, you could of course implement your own comparison logic - for example you could find an open source implementation of XPath deep-equal, and modify it to meet your requirements. It's only a hundred or so lines of code.
Well if you just want to compare and display then you can use Guiffy
It is a good tool. If u want to do the processing in backend then you must use DOM parser load both files to 2 DOM objects and compare attribute by attribute.
it's a bit overkill, but if your XML has schema, you can convert it into EMF metamodel & then use EMF Compare to compare.
