Why should i use url.openStream instead of of url.getContent?

Question

I would like to retrieve the content of a url. Similar to pythons:

html_content = urllib.urlopen("http://www.test.com/test.html").read()

In examples( java2s.com ) you see very often the following code:

URL url = new URL("http://www.test.com/test.html");
String foo = (String) url.getContent();

The Description of getContent is the following:

Gets the contents of this URL. This method is a shorthand for: openConnection().getContent()
Returns: the contents of this URL.

In my opinion that should work perfectly fine. Buuut obviously this code doesnt work, because it raises an error:

Exception in thread "main" java.lang.ClassCastException: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream cannot be cast to java.lang.String

Obviously it returns an inputStream.

So i ask myself: what's the purpose of this function which isn't doing what it is seems to do? And why is no hint for quirks it in the documentation? And why did i saw it in several examples?

Or am i getting this wrong?

The suggested solution (stackoverflow) is to use url.openStream() and then read the Stream.

Florian F · Accepted Answer · 2015-02-22 11:30:11Z

As you said, documentation says that URL.getContent() is a shortcut for openConnection().getContent() so we need to look at the documentation for URLConnection.getContent().

We can see that this returns an Object the type of which is determined by the the content-type header field of the response. This type determines the ContentHandler that will be used. So a ContentHandler converts data based on its MIME type to the appropriate class of Java Object.

In other words the type of Object you get will depend on the content served. For example, it wouldn't make sense to return a String if the MIME type was image/png.

This is why in the example code you link to at java2s.com they check the class of the returned Object:

try {
  URL u = new URL("http://www.java2s.com");
  Object o = u.getContent();
  System.out.println("I got a " + o.getClass().getName());
} catch (Exception ex) {
  System.err.println(ex);
}

So you can say String foo = (String) url.getContent(); if you know your ContentHandler will return a String.

There are default content handlers defined in the sun.net.www.content package but as you can see they are returning streams for you.

You could create your own ContentHandler that does return a String but it will probably be easier just to read the Stream as you suggest.

Martijn Courteaux · Accepted Answer · 2012-03-20 21:43:06Z

3

You misunderstand what "Content" means. You expected it to return a String containing the HTML, but it returns a HttpInputStream. Why? Because the requested URL is a html webpage. Another valid URL might be http://www.google.com/logo.png. This URL doesn't contain String content. It is an image.

edited Mar 20, 2012 at 21:43

answered Mar 20, 2012 at 21:37

Martijn Courteaux

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Comments

prunge · Accepted Answer · 2016-08-30 20:21:49Z

2

You can use Guava's Resources.toString(URL, Charset) method to more easily read a URL to a string.

edited Aug 30, 2016 at 20:21

answered Mar 20, 2012 at 22:09

prunge

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