How to read from an XML within a Zip file in Java

Question

I'm looking to read from a XML file which is stored within a zip file. I'm having TONs of issues understanding how to do this. The best I seem to be able to do is get an input stream working, BUT I CAN'T UNDERSTAND HOW TO READ IT! So I know that its the right file, no I want to pass the data to a XML parser I have. However I don't really understand what the input stream is, or how to manipulate it properly. I've seen it used before with a buffer and then reading into the buffer, my question with this is how do we properly choose the size of the buffer if it is a sufficiently large file?

Any help would be great, thanks!

ZipFile zf;
    try {
        zf = new ZipFile(directory);
        CharBuffer charBuffer = CharBuffer.allocate(BUFFER_SIZE);
      for (Enumeration<? extends ZipEntry> e = zf.entries();
            e.hasMoreElements();) {
        ZipEntry ze = e.nextElement();
        String name = ze.getName();
        if (name.endsWith(".xml")) {
          InputStream in = zf.getInputStream(ze);
          // read from 'in
        }
      }
    } catch (IOException e1) {
        System.out.println("Sorry we couldn't find the file");
        e1.printStackTrace();
    }

Tom · Accepted Answer · 2012-10-07 05:09:26Z

2

You mean as in this?

DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
Document document = docBuilder.parse(in);

Or do you mean you already have a specific XML parser that doesn't take an InputStream - what does it take?

answered Oct 7, 2012 at 5:09

Tom

45.1k4 gold badges44 silver badges61 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Jono Over a year ago

What library is Document from? I'm having some problems with it.

user1406062 Over a year ago

I think org.w3c.dom.Document

KV Prajapati · Accepted Answer · 2012-10-07 05:09:07Z

1

You can either use SAX or StAX to parse/read the stream - Reference JAXP

With Stream API,

 XMLStreamReader reader=XMLInputFactory.newFactory().createXMLStreamReader(in);
 while(reader.hasNext())
 {
  System.out.println(reader.next());
  System.out.println(reader.getLocalName());
 }

With SAX,

SAXParserFactory.newInstance().newSAXParser().parse(in, new YourHandler());

answered Oct 7, 2012 at 5:09

KV Prajapati

94.8k20 gold badges151 silver badges188 bronze badges

Comments

Tom · Accepted Answer · 2012-10-07 05:11:24Z

0

A simple google search can lead you to tons of articles explaining the code work involved to parse XML. Why not just search then?

One of them: http://www.java-samples.com/showtutorial.php?tutorialid=152

I'm not including how to unzip a file because you've mentioned that you're through with that, but if you're looking for options I'll edit my answer.

Happy Java Coding. :)

edited Oct 7, 2012 at 5:11

Tom

45.1k4 gold badges44 silver badges61 bronze badges

answered Oct 7, 2012 at 4:49

Sean Vaughn

3,9421 gold badge19 silver badges14 bronze badges

2 Comments

Jono Over a year ago

XML parsing is the easy part, how do I get my inputStream to convert to an XML? I literally don't know what to do from the inputstream only.

Sean Vaughn Over a year ago

You need to know about the input stream first, in which format is the inputStream? Then study about the format, learn which bits mean what and the basic data parsing approaches for that particular type of data stream. @jQwierdy

Collectives™ on Stack Overflow

How to read from an XML within a Zip file in Java

3 Answers 3

2 Comments

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related