It is possible to read the complete content of a xml files inside a folder with java?

Question

My code is:

DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();      
dbf.setValidating(false);
DocumentBuilder db = dbf.newDocumentBuilder();

Document doc = db.parse(new FileInputStream(new 
File("C:\\Users\\Administrator\\eclipse-workspace\\XMLExamples2\\")));

Element element = (Element) doc.getElementsByTagName("composite").item(0);

// Adds a new attribute. If an attribute with that name is already present 
// in the element, its value is changed to be that of the value parameter
element.setAttribute("xmlns:xsi", "http://www.w3.org/2001/XMLSchema-
instance");
element.setAttribute("xsi:noNamespaceSchemaLocation", "v1_6.xsd");

I'm just able to read and manipulate single files when it put the name after ...\XMLExample2\

Well yes - so loop over all the files in the folder, and load one at a time. — Jon Skeet, Jul 19 '17 at 13:21
Use [File#listFiles()](https://docs.oracle.com/javase/7/docs/api/java/io/File.html#listFiles()) and parse each one — Socratic Phoenix, Jul 19 '17 at 13:21
You have to read the file from the directory in a loop like in the below example. https://stackoverflow.com/questions/3154488/how-do-i-iterate-through-the-files-in-a-directory-in-java — user8271644, Jul 19 '17 at 13:22

Arnaud · Accepted Answer · 2017-07-19T14:00:24.840

1

No, you can't process the content of multiple XML files as a unique document.

But you may list the XML files then process each File from the list :

String xmlDir = "C:\\Users\\Administrator\\eclipse-workspace\\XMLExamples2";
File[] files = new File(xmlDir).listFiles(new FilenameFilter() {

        @Override
        public boolean accept(File folder, String name) {
            return name.toLowerCase().endsWith(".xml");
        }
    });

for(File file : files){

    Document doc = db.parse(file);

    Element element = (Element) doc.getElementsByTagName("composite").item(0);

    // etc...

}

edited Jul 19 '17 at 14:00

answered Jul 19 '17 at 13:23

Arnaud

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i get this error code. Exception in thread "main" java.lang.NullPointerException at AddAttribute_test.main(AddAttribute_test.java:37) line 37 is the first setAttribute – Speedec Jul 19 '17 at 13:30
Try to print the name of the file (`file.getName()`) and check if its content is as expected . – Arnaud Jul 19 '17 at 13:35
you mean in the parse section? Document doc = db.parse(); – Speedec Jul 19 '17 at 13:42
Well the `setAttribute` line shouldn't give you a nullpointer, except if `element` is null. And yes I meant print the file name at the beginning of the loop, to know which one is currently processed . – Arnaud Jul 19 '17 at 13:44
element is null. but why? i defined element in the line above: Element element = (Element) doc.getElementsByTagName("composite").item(0); – Speedec Jul 19 '17 at 13:47
like this? Document doc = db.parse(file.getName()); – Speedec Jul 19 '17 at 13:51
No I just meant to print the name like `System.out.println(file.getName())` so that you know on which file the loop is currently . – Arnaud Jul 19 '17 at 13:52
ah yeah. ofc sorry – Speedec Jul 19 '17 at 13:52
thats wired. he gives me .project back. – Speedec Jul 19 '17 at 13:54
You should filter to get only the XML files, see last edit . – Arnaud Jul 19 '17 at 14:00
ok the Filefilter is working, but access is denied to the other .xml files. – Speedec Jul 19 '17 at 14:07

score -1 · Answer 2 · answered Jul 19 '17 at 13:26

Each file is a seperate document. So, if you aren't trying to read them as a single document, then all you need to do is grab a list of files in that folder using the File class' listFiles() method. You can then loop over each file in the folder and process them accordingly.

If you do want to treat it as a single XML document: you will need to merge the files together first, and then process it. There are a number of ways that can be done. You may want to look at this question for a couple ideas.

It is possible to read the complete content of a xml files inside a folder with java?

2 Answers2