I´d need to forward a url like
http://localhost:8080/doc/service?type=testLogin&exectype=userShow&moreparms
to:
http://localhost:8080/xyz/service?type=testLogin&exectype=userShow&moreparms
The system runs with tomcat 8.0.28 and has already the URLRewriteFilter (http://www.tuckey.org/urlrewrite/) loaded, but is not working with the follow specific code that i mentioned on the urlrewritefilter-site at googlecode.com that needs to be put in the webapps/doc/WEB-INF/web.xml:
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5">
<filter>
<filter-name>UrlRewriteFilter</filter-name>
<filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
<init-param>
<param-name>logLevel</param-name>
<param-value>DEBUG</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>UrlRewriteFilter</filter-name>
<url-pattern>/*</url-pattern>
<dispatcher>REQUEST</dispatcher>
<dispatcher>FORWARD</dispatcher>
</filter-mapping>
And i added the follow rule to urlrewrite.xml in the same directory as web.xml:
<urlrewrite>
<rule>
<from>^/doc/(.*)$</from>
<to type="redirect">/xyz/$1</to>
</rule>
<outbound-rule>
</outbound-rule>
<urlrewrite>
I'm getting 404's when trying to go to something like
http://localhost:8080/doc/service?type=testLogin&exectype=userShow&moreparms
I get the 404 error also if i try
http://localhost:8080/doc/status
with the example-rule like:
<urlrewrite>
<rule>
<from>/doc/status/</from>
<to type="redirect">%{context-path}/rewrite-status</to>
</rule>
<outbound-rule>
<from>/rewrite-status</from>
<to>/doc/status/</to>
</outbound-rule>
<urlrewrite>
But
http://localhost:8080/doc/rewrite-status
works.
Where is my fault?
P.S: I`m using Win10 pro and urlrewritefilter-4.0.3