IOException - XML Parsing

Question

I am trying to parse an xml response from the below url -

http://imdbapi.org/?type=xml&q=argo

For this, i have written the below code -

    try
    {
        XMLReader myReader = XMLReaderFactory.createXMLReader();
        xmlHandler handlerobj = new xmlHandler();
        myReader.setContentHandler(handlerobj);
        myReader.parse(new InputSource(new URL("http://imdbapi.org/?type=xml&q=argo").openStream()));
    }
    catch(Exception e)
    {
        System.out.println("Error");
    }

xmlHandler is a class that extends DefaultHandler. I am getting an IOException in the above code.

Stack trace -

java.io.IOException: Server returned HTTP response code: 403 for URL:   http://imdbapi.org/?type=xml&q=argo
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(Unknown Source)
at java.net.URL.openStream(Unknown Source)
at gui.getimdbdata(gui.java:73)
at gui.main(gui.java:64)

What is the problem with this code ?

The server refuses to allow the requested action (response code: 403) Everything is ok, with server configuration? Are you able to get the xml file in other ways? — Quirin
– Quirin, Commented Jun 6, 2013 at 7:22
I have a similar android project which does the same, parses the xml response from this same URL. I can see the xml using the browser too. — Ankit Rustagi
– Ankit Rustagi, Commented Jun 6, 2013 at 7:24
This url acceps only requests from browser. You can find solution here: stackoverflow.com/questions/4797593/… — dijkstra
– dijkstra, Commented Jun 6, 2013 at 7:25

ruben056 · Accepted Answer · 2013-06-06 08:01:12Z

2

You must set the user.agent:

System.setProperty("http.agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36     (KHTML, like Gecko) Chrome/28.0.1500.29 Safari/537.36");

(if you connect to the URL with your browser this is done automagically)

answered Jun 6, 2013 at 8:01

ruben056

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Ankit Rustagi · Accepted Answer · 2013-06-06 08:51:34Z

0

Solved the issue, thanks to @dijkstra !

The web service would only allow browser to fetch the xml data.

Following are the modifications -

       url = new URL(urlString);
       uc = url.openConnection();
       uc.addRequestProperty("User-Agent", 
                "Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.0)");

       uc.connect();
       uc.getInputStream();
       BufferedInputStream in = new BufferedInputStream(uc.getInputStream());

answered Jun 6, 2013 at 8:51

Ankit Rustagi

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