Java - Convert String to valid URI object

lostInTransit_46297 0
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I am trying to get a java.net.URI object from a String. The string has some characters which will need to be replaced by their percentage escape sequences. But when I use URLEncoder to encode the String with UTF-8 encoding, even the / are replaced with their escape sequences.

How can I get a valid encoded URL from a String object?

http://www.google.com?q=a b gives http%3A%2F%2www.google.com... whereas I want the output to be http://www.google.com?q=a%20b

Can someone please tell me how to achieve this.

I am trying to do this in an Android app. So I have access to a limited number of libraries.

utf-8 encoding android java
11 Answers

You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project

Like this (see URIUtil):

URIUtil.encodeQuery("http://www.google.com?q=a b")

will become:

http://www.google.com?q=a%20b

You can of course do it yourself, but URI parsing can get pretty messy...

Hans_Doggen_95040

You can use the multi-argument constructors of the URI class. From the URI javadoc:

The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.

So if you use

URI uri = new URI("http", "www.google.com?q=a b");

Then you get http:www.google.com?q=a%20b which isn't quite right, but it's a little closer.

If you know that your string will not have URL fragments (e.g. http://example.com/page#anchor), then you can use the following code to get what you want:

String s = "http://www.google.com?q=a b";
String[] parts = s.split(":",2);
URI uri = new URI(parts[0], parts[1], null);

To be safe, you should scan the string for # characters, but this should get you started.

user_7370
TofuBeer_658680

Or perhaps you could use this class:

http://developer.android.com/reference/java/net/URLEncoder.html

Which is present in Android since API level 1.

Annoyingly however, it treats spaces specially (replacing them with + instead of %20). To get round this we simply use this fragment:

URLEncoder.encode(value, "UTF-8").replace("+", "%20");

user_684300
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