-Tim
Bill Faulk wrote:
The same error occurs for "%25". If I put a "%" anywhere in the
parameter I get a crash. "4%" would pass a parameter to query everything
starting with 4 in my example.
--- Bill Faulk [EMAIL PROTECTED] 801-302-8930
-----Original Message-----
From: Tim Funk [mailto:[EMAIL PROTECTED] Sent: Monday, February 16, 2004 11:12 AM
To: Tomcat Users List
Subject: Re: request.getParameter Error when "%" contained in parameter
value
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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