URL encoding in Android


Question

How do you encode a URL in Android?

I thought it was like this:

final String encodedURL = URLEncoder.encode(urlAsString, "UTF-8");
URL url = new URL(encodedURL);

If I do the above, the http:// in urlAsString is replaced by http%3A%2F%2F in encodedURL and then I get a java.net.MalformedURLException when I use the URL.

1
316
5/25/2015 7:39:41 PM

Accepted Answer

You don't encode the entire URL, only parts of it that come from "unreliable sources".

String query = URLEncoder.encode("apples oranges", "utf-8");
String url = "http://stackoverflow.com/search?q=" + query;

Alternatively, you can use Strings.urlEncode(String str) of DroidParts that doesn't throw checked exceptions.

Or use something like

String uri = Uri.parse("http://...")
                .buildUpon()
                .appendQueryParameter("key", "val")
                .build().toString();
609
10/4/2013 3:32:22 PM

I'm going to add one suggestion here. You can do this which avoids having to get any external libraries.

Give this a try:

String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();

You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.

This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.

The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.


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