Java: url encoding leaving 'allowed' character intact

Question

Simple question from a Java novice. I want to encode a url so that nonstandard characters will be transformed to their hex value (that is %XX) while characters one expects to see in a url - letter, digits, forward slashes, question marks and whatever, will be left intact.

For example, encoding

"hi/hello?who=moris\\boris"

should result with

"hi/hello?who=moris%5cboris"

ideas?

Answer 1

This is actually, a rather tricky problem. And the reason that it is tricky is that the different parts of a URL need to be handled (encoded) differently.

In my experience, the best way to do this is to assemble the url from its components using the URL or URI class, letting the them take care of the encoding the components correctly.

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In fact, now that I think about it, you have to encode the components before they get assembled. Once the parts are assembled it is impossible to tell whether (for example) a "?" is intended to the query separator (don't escape it) or a character in a pathname component (escape it).