Encoding URL query parameters in Java

问题

How does one encode query parameters to go on a url in Java? I know, this seems like an obvious and already asked question.

There are two subtleties I\'m not sure of:

Should spaces be encoded on the url as \"+\" or as \"%20\"? In chrome if I type in \"http://google.com/foo=?bar me\" chrome changes it to be encoded with %20
Is it necessary/correct to encode colons \":\" as %3B? Chrome doesn\'t.

Notes:

java.net.URLEncoder.encode doesn\'t seem to work, it seems to be for encoding data to be form submitted. For example, it encodes space as + instead of %20, and encodes colon which isn\'t necessary.
java.net.URI doesn\'t encode query parameters

回答1:

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.

回答2:

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.

回答3:

Unfortunately, URLEncoder.encode() does not produce valid percent-encoding (as specified in http://tools.ietf.org/html/rfc3986#section-2.1).

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);

回答4:

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.

回答5:

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");

回答6:

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

来源：https://stackoverflow.com/questions/5330104/encoding-url-query-parameters-in-java

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java

urlencode