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From Tim Funk <funk...@joedog.org>
Subject Re: request.getParameter Error when "%" contained in parameter value
Date Mon, 16 Feb 2004 18:12:18 GMT
Why don't you use: http://website/findnaics.jsp?code=%25

-Tim

Bill Faulk wrote:

> I am using Tomcat 4.1.29 on Windows 2000 Server with Java SDK 1.4.2.
> 
> On the user query forms in my application "%" is a valid wildcard for
> "all" as it is in the actual database query. I don't want to use blank
> for all because I don't want users to accidentally search for all
> records; i.e. they have to actually enter % to search for all records in
> a value. Blank parameters are ignored. 
> 
> Passing % as a parameter via GET or POST causes the error
> 
> http://website/findnaics.jsp?code=4%
> 
> I am using POST methods for the forms as in...
> 
> <form method="POST" name="findform" action="findnaics.jsp">
> 
> The request.getParameter line is actually generating the error if the
> parameter contains a %.
> 
> String code = request.getParameter("code");
> 
> I've seen this error when searching in regards to forwarding pages
> (http://nagoya.apache.org/bugzilla/show_bug.cgi?id=3986) but it was
> considered "invalid" because forward expects an encoded url. However, I
> am simply passing a parameter and submitting a form. Doing something
> like 'action = <%= response.encodeURL("findnaics.jsp") %>' doesn't do
> anything for me. Using quotes/backslash, etc. doesn't make any
> difference. These query parameters can be passed by both GET and POST so
> encoding the URL isn't an option even if it did work.
> 
> When "%" is passed as a parameter I get the following error for the
> getParameter statement.
> 
> java.io.CharConversionException: EOF
>         at
> org.apache.tomcat.util.buf.UDecoder.convert(UDecoder.java:177)



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