Posted By: Anonymous
Say I have a URL
and I have a query entered by the user such as:
random word £500 bank $
I want the result to be a properly encoded URL:
What’s the best way to achieve this? I tried
URLEncoder and creating URI/URL objects but none of them come out quite right.
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
Note that spaces in query parameters are represented by
%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.