Encoding URL query parameters in Java

java.net.URLEncoder.encode(String s, String encoding) can help too. It follows the HTML form encoding application/x-www-form-urlencoded.

URLEncoder.encode(query, "UTF-8");

On the other hand, Percent-encoding (also known as URL encoding) encodes space with %20. Colon is a reserved character, so : will still remain a colon, after encoding.

Unfortunately, URLEncoder.encode() does not produce valid percent-encoding (as specified in RFC 3986).

URLEncoder.encode() encodes everything just fine, except space is encoded to "+". All the Java URI encoders that I could find only expose public methods to encode the query, fragment, path parts etc. - but don't expose the "raw" encoding. This is unfortunate as fragment and query are allowed to encode space to +, so we don't want to use them. Path is encoded properly but is "normalized" first so we can't use it for 'generic' encoding either.

Best solution I could come up with:

return URLEncoder.encode(raw, "UTF-8").replaceAll("\\+", "%20");

If replaceAll() is too slow for you, I guess the alternative is to roll your own encoder...

EDIT: I had this code in here first which doesn't encode "?", "&", "=" properly:

//don't use - doesn't properly encode "?", "&", "="
new URI(null, null, null, raw, null).toString().substring(1);

EDIT: URIUtil is no longer available in more recent versions, better answer at Java - encode URL or by Mr. Sindi in this thread.

URIUtil of Apache httpclient is really useful, although there are some alternatives

URIUtil.encodeQuery(url);

For example, it encodes space as "+" instead of "%20"

Both are perfectly valid in the right context. Although if you really preferred you could issue a string replace.

It is not necessary to encode a colon as %3B in the query, although doing so is not illegal.

URI         = scheme ":" hier-part [ "?" query ] [ "#" fragment ]
query       = *( pchar / "/" / "?" )
pchar         = unreserved / pct-encoded / sub-delims / ":" / "@"
unreserved    = ALPHA / DIGIT / "-" / "." / "_" / "~"
pct-encoded   = "%" HEXDIG HEXDIG
sub-delims    = "!" / "$" / "&" / "'" / "(" / ")" / "*" / "+" / "," / ";" / "="

It also seems that only percent-encoded spaces are valid, as I doubt that space is an ALPHA or a DIGIT

look to the URI specification for more details.

The built in Java URLEncoder is doing what it's supposed to, and you should use it.

A "+" or "%20" are both valid replacements for a space character in a URL. Either one will work.

A ":" should be encoded, as it's a separator character. i.e. http://foo or ftp://bar. The fact that a particular browser can handle it when it's not encoded doesn't make it correct. You should encode them.

As a matter of good practice, be sure to use the method that takes a character encoding parameter. UTF-8 is generally used there, but you should supply it explicitly.

URLEncoder.encode(yourUrl, "UTF-8");

I just want to add anther way to resolve this problem.

If your project depends on spring web, you can use their utils.

import org.springframework.web.util.UriUtils

import java.nio.charset.StandardCharsets

UriUtils.encode('vip:104534049:5', StandardCharsets.UTF_8)

Output:

vip%3A104534049%3A5

String param="2019-07-18 19:29:37";
param="%27"+param.trim().replace(" ", "%20")+"%27";

I observed in case of Datetime (Timestamp) URLEncoder.encode(param,"UTF-8") does not work.

The white space character " " is converted into a + sign when using URLEncoder.encode. This is opposite to other programming languages like JavaScript which encodes the space character into %20. But it is completely valid as the spaces in query string parameters are represented by +, and not %20. The %20 is generally used to represent spaces in URI itself (the URL part before ?).

if you have only space problem in url. I have used below code and it work fine

String url;
URL myUrl = new URL(url.replace(" ","%20"));

example : url is

www.xyz.com?para=hello sir

then output of muUrl is

www.xyz.com?para=hello%20sir