Home/ Questions/Q 359747
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-12T12:25:23+00:00

i am getting adding some of parameter in query string.value of these param can

  • 0

i am getting adding some of parameter in query string.value of these param can be “a%%”,”%” etc.on java code side .while parsing query parameter i m getting char conversionexception
as shown below in exception log.

13:14:39,555 ERROR [STDERR] java.io.CharConversionException: EOF
13:14:39,555 ERROR [STDERR] at org.apache.tomcat.util.buf.UDecoder.convert(UDecoder.java:119)
13:14:39,555 ERROR [STDERR] at org.apache.tomcat.util.buf.UDecoder.convert(UDecoder.java:87)
13:14:39,555 ERROR [STDERR] at org.apache.tomcat.util.http.Parameters.processParameters(Paramete
rs.java:428)
13:14:39,555 ERROR [STDERR] at org.apache.tomcat.util.http.Parameters.processParameters(Paramete
rs.java:515)
13:14:39,555 ERROR [STDERR] at org.apache.tomcat.util.http.Parameters.handleQueryParameters(Para
meters.java:298)
:14:39,555 ERROR [STDERR] at org.apache.coyote.tomcat4.CoyoteRequest.parseRequestParameters(Co
yoteRequest.java:1933)

what should i do?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    It sounds like the query string being sent to your Java code is malformed. The percent sign is special in query strings: It introduces a two-digit hexadecimal number identifying a character. For instance, %20 is a space. To put a percent sign in a query string correctly, one uses %25 (character 25h is the percent sign in Unicode). If the query string you’re processing really, literally has %% in it, then it is malformed and you’ll want to have the side sending it fixed.

    Edit: In your comment you say you’re the one sending the invalid query string. To properly encode a query parameter, use the encodeURIComponent JavaScript function:

    var encodedValue;
encodedValue = encodeURIComponent(yourTextfield.value);
someurl = "http://example.com?x=" + encodedValue;
    • 0