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Encoding URL query parameters in Java

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java

urlencode

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How do you encode a URL in Java?

Encode the URLprivate String encodeValue(String value) { return URLEncoder. encode(value, StandardCharsets. UTF_8. toString()); } @Test public void givenRequestParam_whenUTF8Scheme_thenEncode() throws Exception { Map<String, String> requestParams = new HashMap<>(); requestParams.

Should query parameters be URL encoded?

Why do we need to encode? URLs can only have certain characters from the standard 128 character ASCII set. Reserved characters that do not belong to this set must be encoded. This means that we need to encode these characters when passing them into a URL.


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.


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.


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);

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");