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Validation Error while deploying war file in Weblogic

I got the following error while deploying war file in weblogic
Message icon - Error VALIDATION PROBLEMS WERE FOUND <5:4> problem: cvc->complex-type.2.3: Element 'web-app#http://java.sun.com/xml/ns/javaee' with element-only content type cannot have text content. <14:5> problem: cvc-complex-type.2.4a: Expected element 'url-pattern#http://java.sun.com/xml/ns/javaee' instead of 'load-on-startup#http://java.sun.com/xml/ns/javaee' here in element servlet-mapping#http://java.sun.com/xml/ns/javaee <18:5> problem: cvc-complex-type.2.4b: Element not allowed: listener-class#http://java.sun.com/xml/ns/javaee in element listener#http://java.sun.com/xml/ns/javaee
Below is my web.xml
<web-app id="WebApp_ID" version="2.4"
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/j2ee ">
>
<display-name>Archetype Created Web Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
<load-on-startup>1</load-on-startup>
</servlet-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
<listener-class>
org.springframework.security.web.session.HttpSessionEventPublisher
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/mvc-dispatcher-servlet.xml
</param-value>
</context-param>
and here's my weblogic.xml
<?xml version="1.0" encoding="UTF-8"?>
<weblogic-web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.oracle.com/weblogic/weblogic-web-app"
xsi:schemaLocation="http://xmlns.oracle.com/weblogic/weblogic-web-app ht
tp://xmlns.oracle.com/weblogic/weblogic-web-app/1.4/weblogic-web-app.xsd">
<context-root>/</context-root>
<container-descriptor>
<show-archived-real-path-enabled>true</show-archived-real-path-enabled>
<prefer-application-packages>
<package-name>javax.faces.*</package-name>
<package-name>com.sun.faces.*</package-name>
<package-name>com.bea.faces.*</package-name>
<package-name>org.slf4j.*</package-name>
<package-name>com.google.common.*</package-name>
</prefer-application-packages>
<prefer-application-resources>
<resource-name>javax.faces.*</resource-name>
<resource-name>com.sun.faces.*</resource-name>
<resource-name>com.bea.faces.*</resource-name>
<resource-name>com.google.common.*</resource-name>
<resource-name>org/slf4j/impl/StaticLoggerBinder.class</resource-name>
<resource-name>META-INF/services/javax.servlet.ServletContainerInitializer</resource-name>
</prefer-application-resources>
</container-descriptor>
<session-descriptor>
<cookie-secure>true</cookie-secure>
<url-rewriting-enabled>false</url-rewriting-enabled>
<!-- <cookie-http-only>true</cookie-http-only> -->
</session-descriptor>
<virtual-directory-mapping>
<local-path>/usr/local/WebLogic/wlApps/miDocs</local-path>
<url-pattern>/docs/*</url-pattern>
</virtual-directory-mapping>
The folder structure matches as per the description in the oracle website
Any help is appreciated

Thinky ou have two Problems here:
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee ">
>
First a blank at the end of the schemaLocation.
Second > is too much

<web-app id="WebApp_ID" version="2.4"
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/j2ee ">
>
<display-name>Archetype Created Web Application</display-name>
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<listener>
<listener-class>
org.springframework.security.web.session.HttpSessionEventPublisher
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/mvc-dispatcher-servlet.xml
</param-value>
</context-param>
</web-app>

How to redirect from one servlet to another? [duplicate]

This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 5 years ago.
The xml file is located in WebContent/WEB-INF/web.xml of my project. I am using Eclipse and running Tomcat (which is not installed via Eclipse. I prefer it to be a separate installation).
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>EmployeeManagement</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
It doesnt work when the form page submits to the servlet. I am getting a 404 error everytime. I have been encountering this problem for a while. Somebody please help me.

You are missing <servlet>...</servlet> tag which is important to mapping. So use following :
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>EmployeeManagement</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
and you should give action value on your form like following:
<form action="/EmployeeManagement/WebContent/Registration" method="post">
//Some code here
</form>
and also note it down all values are case sensitive on the following code:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Your servlet name Registration should be same on both tags <servlet>...</servlet> and <servlet-mapping>...</servlet-mapping> and also package name should be same where your servlet class is located.

you have not mapped servlet name to servlet class, It should be like given below,
In <servlet-class> give the path of your servlet
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.Registration<servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>

Check your form action.
Is the path there
/EmployeeManagement/WebContent/Registration
or
YOURAPPCONTEXT/EmployeeManagement/WebContent/Registration
or
YOURAPPNAME/EmployeeManagement/WebContent/Registration

You have specifyed a servlet-mapping and used the name Registration in servlet-name without defining it before.
You need to define the servlet before using it in a servlet mapping
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>[fully qualifyied name of your servlet]</servlet-class>
</servlet>

You are missing another part to define the servlet in the web.xml
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>
package.path.to.RegistrationServlet
</servlet-class>
</servlet>

You forgot a vital part of the configuration. You should add this to your web.xml before servlet-mapping tag:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.name.of.your.servlet.class</servlet-class>
</servlet>

Rest Easy via html is not working

I have a rest easy url. When i type the url in browser java class get called returns the output. when i call the index.html file it is not loading. i am using tomcat7
This is my web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com /xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns
/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>projectname</display-name>
<resource-ref>
<description>MySQL Datasource example</description>
<res-ref-name>jdbc/projectnamedb</res-ref-name>
<res-type>javax.sql.DataSource</res-type>
<res-auth>Container</res-auth>
</resource-ref>
<context-param>
<param-name>log4jConfigLocation</param-name>
<param-value>${catalina.home}\conf\log4j.properties</param-value>
</context-param>
<context-param>
<param-name>inputPropertiesLocation</param-name>
<param-value>${catalina.home}\conf\projectDetails.properties</param-value>
</context-param>
<context-param>
<param-name>resteasy.scan</param-name>
<param-value>true</param-value>
</context-param>
<listener>
<listener-class>org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap
</listener-class>
</listener>
<servlet>
<servlet-name>Resteasy</servlet-name>
<servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher
</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Resteasy</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>/index.html</welcome-file>
</welcome-file-list>
</web-app>

Well 2 things to know here
1) You cannot type simply http://index.html you have to give path to the index page like http://projectname/path/to/index.html
2) The other simple way is to add index.html as welcome page. So when-ever server is started it opens index.html page at first. like
<welcome-file-list>
<welcome-file>path/to/index.html</welcome-file>
</welcome-file-list>

Mapping servlet in web.xml [duplicate]

This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 5 years ago.
The xml file is located in WebContent/WEB-INF/web.xml of my project. I am using Eclipse and running Tomcat (which is not installed via Eclipse. I prefer it to be a separate installation).
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>EmployeeManagement</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
It doesnt work when the form page submits to the servlet. I am getting a 404 error everytime. I have been encountering this problem for a while. Somebody please help me.

You are missing <servlet>...</servlet> tag which is important to mapping. So use following :
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>EmployeeManagement</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
and you should give action value on your form like following:
<form action="/EmployeeManagement/WebContent/Registration" method="post">
//Some code here
</form>
and also note it down all values are case sensitive on the following code:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Your servlet name Registration should be same on both tags <servlet>...</servlet> and <servlet-mapping>...</servlet-mapping> and also package name should be same where your servlet class is located.

you have not mapped servlet name to servlet class, It should be like given below,
In <servlet-class> give the path of your servlet
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.Registration<servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>

Check your form action.
Is the path there
/EmployeeManagement/WebContent/Registration
or
YOURAPPCONTEXT/EmployeeManagement/WebContent/Registration
or
YOURAPPNAME/EmployeeManagement/WebContent/Registration

You have specifyed a servlet-mapping and used the name Registration in servlet-name without defining it before.
You need to define the servlet before using it in a servlet mapping
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>[fully qualifyied name of your servlet]</servlet-class>
</servlet>

You are missing another part to define the servlet in the web.xml
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>
package.path.to.RegistrationServlet
</servlet-class>
</servlet>

You forgot a vital part of the configuration. You should add this to your web.xml before servlet-mapping tag:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.name.of.your.servlet.class</servlet-class>
</servlet>

Cause of "Error deploying configuration descriptor" on Tomcat 7.0.12

When I try to deploy a WAR file by dropping it into webapps directory, I get following message in the console:
04.05.2011 19:34:07 org.apache.catalina.startup.HostConfig deployWAR
SEVERE: Error deploying configuration descriptor xyz.war
I didn't find any hints about the detailed cause of this in the logs.
Where can I find out the details (e. g. the stack trace) about this failure?
Here's the web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID" version="2.5">
<display-name>myapp</display-name>
<context-param>
<description>
Vaadin production mode</description>
<param-name>productionMode</param-name>
<param-value>false</param-value>
</context-param>
<servlet>
<servlet-name>Project Control Center Application</servlet-name>
<servlet-class>com.vaadin.terminal.gwt.server.ApplicationServlet</servlet-class>
<init-param>
<description>
Vaadin application class to start</description>
<param-name>application</param-name>
<param-value>at.mycompany.myapp.ProjectControlCenterApplication</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Project Control Center Application</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<listener>
<listener-class>at.mycompany.myapp.impl.persistence.DatabaseStartStopServletContextListener</listener-class>
</listener>
<session-config>
<session-timeout>480</session-timeout>
</session-config>
</web-app>

I found the answer. The cause of the problem was a malformed context.xml file.

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" ...>
<welcome-file-list>
<welcome-file>
...
</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>info</servlet-name>
<servlet-class>servlets.info.Properties</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Properties</servlet-name>
<!-- Here I forget to change the servlet name to info -->
<!-- So, the reason is invalid web.xml file -->
<url-pattern>/Properties</url-pattern>
</servlet-mapping>
</web-app>
Double-check your web.xml file.