[javascript] How to encode URL parameters?

Just try encodeURI() and encodeURIComponent() yourself...

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console.log(encodeURIComponent('@#$%^&*'));
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Input: @#$%^&*. Output: %40%23%24%25%5E%26*. So, wait, what happened to *? Why wasn't this converted? TLDR: You actually want fixedEncodeURIComponent() and fixedEncodeURI(). Long-story...

You should not be using encodeURIComponent() or encodeURI(). You should use fixedEncodeURIComponent() and fixedEncodeURI(), according to the MDN Documentation.

Regarding encodeURI()...

If one wishes to follow the more recent RFC3986 for URLs, which makes square brackets reserved (for IPv6) and thus not encoded when forming something which could be part of a URL (such as a host), the following code snippet may help:

function fixedEncodeURI(str) { return encodeURI(str).replace(/%5B/g, '[').replace(/%5D/g, ']'); }

Regarding encodeURIComponent()...

To be more stringent in adhering to RFC 3986 (which reserves !, ', (, ), and *), even though these characters have no formalized URI delimiting uses, the following can be safely used:

function fixedEncodeURIComponent(str) { return encodeURIComponent(str).replace(/[!'()*]/g, function(c) { return '%' + c.charCodeAt(0).toString(16); }); }

So, what is the difference? fixedEncodeURI() and fixedEncodeURIComponent() convert the same set of values, but fixedEncodeURIComponent() also converts this set: +@?=:*#;,$&. This set is used in GET parameters (&, +, etc.), anchor tags (#), wildcard tags (*), email/username parts (@), etc..

For example -- If you use encodeURI(), [email protected]/?email=me@home will not properly send the second @ to the server, except for your browser handling the compatibility (as Chrome naturally does often).