Sorry My web.xml was in wrong place. Now I cannot start my app. It says ERROR: ListenerStart. Why is this hyappening?
On 2/14/08, Arun <[EMAIL PROTECTED]> wrote: > > Hi, > > I had been following the example in > http://struts.apache.org/2.x/docs/struts-2-spring-2-jpa-ajax.html to build > my own app. > But I do not want to use the custom library or tiles or ognl or struts > tags or dojo. I am using plain jsp and el. > I just want to hit the POJO action class thru url. > What url should I use? > > Is it needed that I use s:url. If not how I can construct my url . I tried > http://localhost:8080/packagename/actionname.action > and http://localhost:8080/<executeactionname>.action?method=methodname > > Both did not work in my case. I got a 404 error. > I checked that struts.xml is in classpath and I have a persistence.xml in > META-INF directory too. > And also I have an applicationContext.xml in WEB-INF too. > > My struts.xml looks like > > <?xml version="1.0" encoding="UTF-8" ?> > <!DOCTYPE struts PUBLIC > "-//Apache Software Foundation//DTD Struts Configuration 2.0//EN" > "http://struts.apache.org/dtds/struts-2.0.dtd";> > <struts> > <constant name="struts.objectFactory" value="spring" /> > <constant name="struts.devMode" value="true" /> > > <package name="employeeregistration" extends="struts-default" > > > <action name="employeeAcct" method="execute" > class="employeeAction"> > <result>/index.jsp</result> > <result name="input">/index.jsp</result> > </action> > <action name="createAccount" method="createEmployeeAccount" > class="employeeAction"> > <result>/index.jsp</result> > <result name="input">/index.jsp</result> > </action> > > > </package> > > </struts> > > > > > > > -- > Thanks > Arun George -- Thanks Arun George
