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This question already has an answer here:

As the title says: which encoder would give me space as %20 as opposed to +? I need it for android. java.net.URLEncoder.encode gives +

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marked as duplicate by Ted Hopp, morgano, laalto, Shankar Damodaran, Charles Caldwell Aug 20 at 17:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3 Answers

up vote 5 down vote

Android has it's own Uri class which you could use.

E.g.

String url = Uri.parse("http://www.google.com").buildUpon()
    .appendQueryParameter("q", "foo bar")
    .appendQueryParameter("xml", "<Hellö>")
    .build().toString();

results in

http://www.google.com?q=foo%20bar&xml=%3CHell%C3%B6%3E

Uri Encodes characters in the given string as '%'-escaped octets using the UTF-8 scheme. Leaves letters ("A-Z", "a-z"), numbers ("0-9"), and unreserved characters ("_-!.~'()*") intact.

Note: only _-.* are considered unreserved characters by URLEncoder. !~'() would get converted to %21%7E%27%28%29.

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up vote 0 down vote

You have to replace the + by yourself.

Example: 

System.out.println(java.net.URLEncoder.encode("Hello World", "UTF-8").replace("+", "%20"));

For more look at this post:

URLEncoder not able to translate space character

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up vote 0 down vote

As mentioned here, you could just replace the string yourself using 

String temp = "a+url+with+plus+signs";

temp = temp.replaceAll("+", "%20");
URL sourceUrl = new URL(temp);
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This is wrong. Other special characters present in the text may also need to be escaped. If you escape the space characters first like that and then use URLEncoder.encode, the % in %20 will itself be encoded as %25. If you don't use URLEncoder.encode, then you risk other special characters remaining in the text. Much better to encode first and then replace the + with %20. –  Ted Hopp Aug 19 at 21:26
 
Good point. I will update my answer. –  Adam Johns Aug 19 at 21:29
 
You probably should just use temp.replace("+", "%20"); replaceAll expects a regex as the first parameter and the + would need to be escaped as "\\+". –  Ted Hopp Aug 19 at 21:36

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