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Escaping & in a URL

开发者 https://www.devze.com 2022-12-19 16:54 出处:网络
I am using jsps and in my url I have a value for a variable l开发者_开发技巧ike say \"L & T\". Now when I try to retrieve the value for it by using request.getParameter I get only \"L\". It recogn

I am using jsps and in my url I have a value for a variable l开发者_开发技巧ike say "L & T". Now when I try to retrieve the value for it by using request.getParameter I get only "L". It recognizes "&" as a separator and thus it is not getting considered as a whole string.

How do I solve this problem?


java.net.URLEncoder.encode("L & T", "utf8")

this outputs the URL-encoded, which is fine as a GET parameter:

L+%26+T


A literal ampersand in a URL should be encoded as: %26

// Your URL
http://www.example.com?a=l&t

// Encoded
http://www.example.com?a=l%26t


You need to "URL encode" the parameters to avoid this problem. The format of the URL query string is: ...?<name>=<value>&<name>=<value>&<etc> All <name>s and <value>s need to be URL encoded, which basically means transforming all the characters that could be interpreted wrongly (like the &) into %-escaped values. See this page for more information: http://www.w3schools.com/TAGS/ref_urlencode.asp

If you're generating the problem URL with Java, you use this method: String str = URLEncoder.encode(input, "UTF-8");

Generating the URL elsewhere (some templates or JS or raw markup), you need to fix the problem at the source.


You can use UriUtils#encode(String source, String encoding) from Spring Web. This utility class also provides means for encoding only some parts of the URL, like UriUtils#encodePath.

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