Java URL encoding of query string parameters
URLEncoder
is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character &
nor the parameter name-value separator character =
.
String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);
When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString()
as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8"
.
Note that spaces in query parameters are represented by +
, not %20
, which is legitimately valid. The %20
is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?
), not in query string (the part after ?
).
Also note that there are three encode()
methods. One without Charset
as second argument and another with String
as second argument which throws a checked exception. The one without Charset
argument is deprecated. Never use it and always specify the Charset
argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.
All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
See also:
- What every web developer must know about URL encoding
I would not use URLEncoder
. Besides being incorrectly named (URLEncoder
has nothing to do with URLs), inefficient (it uses a StringBuffer
instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.
Instead I would use URIBuilder
or Spring's org.springframework.web.util.UriUtils.encodeQuery
or Commons Apache HttpClient
.
The reason being you have to escape the query parameters name (ie BalusC's answer q
) differently than the parameter value.
The only downside to the above (that I found out painfully) is that URL's are not a true subset of URI's.
Sample code:
import org.apache.http.client.utils.URIBuilder;
URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();
// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24
Since I'm just linking to other answers I marked this as a community wiki. Feel free to edit.