Validation Error while deploying war file in Weblogic - java

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>

Related

Can we replace word contextConfigLocation by foo?

I am using Java/JDK 1.6, Spring 2.5.6, Eclipse IDE 2022-06 . I see file web.xml
<web-app xmlns="http://java.sun.com/xml/ns/javaee" 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">
<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>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/mvc-dispatcher-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
Please focus at this line
<param-name>contextConfigLocation</param-name>
Can we replace word contextConfigLocation by foo?

Jersey + spring welcome file

I have index.html in my WEB-INF folder. I want it to be loaded when I go to localhost:8080/myapp. But unfortunately I am getting 404.
Could you please let me know what do I need to modify in order achieve that?
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<display-name>${project.name} (${project.version})</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<servlet>
<servlet-name>jersey-servlet</servlet-name>
<servlet-class>
com.sun.jersey.spi.spring.container.servlet.SpringServlet
</servlet-class>
<init-param>
<param-name>
com.sun.jersey.config.property.packages
</param-name>
<param-value>com.skygraph.facade</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-servlet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>

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>

Spring configuration not finding with Configured name [duplicate]

This question already has answers here:
Why does Spring MVC respond with a 404 and report "No mapping found for HTTP request with URI [...] in DispatcherServlet"?
(13 answers)
Closed 6 years ago.
The following are my configuration file i configured
WEB-INF/classes/applicationContext.xml but it searching for
WEB-INF/applicationContext.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="starter_anil" 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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Anil-Spring</display-name>
<!-- Servlets -->
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>WEB-INF/classes/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<!-- Listeners -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
i am getting following exception
EVERE: Context initialization failed
org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/applicationContext.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/applicationContext.xml]
here is my applicationContext
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">
<beans>
<import resource="config/controllers.xml" />
<!-- <import resource="conf/spring/services.xml" />
<import resource="conf/spring/persistence.xml" />
<import resource="conf/spring/daos.xml" />
<import resource="conf/spring//interceptors.xml" /> -->
</beans>
the solution for this is
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="starter_anil" 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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Anil-Spring</display-name>
<!-- Servlets -->
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</context-param>
<!-- Listeners -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
WARNING: No mapping found for HTTP request with URI
[/AnilsSpring/WEB-INF/jsp/hello.jsp] in DispatcherServlet with name
'mvc-dispatcher'
please guide line me once

Why is your applicationContext is in the classes? Move it to the WEB-INF directory, as it is told in the comment. That should fix the problem.
The /classes directory is for the compiled classes, but not for the configuration files.

Try adding a / in /WEB-INF/classes/applicationContext.xml
If it still doesn't work, try adding this instead in your web.xml (before the servlet declaration)
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/classes/applicationContext.xml
</param-value>
</context-param>

<?xml version="1.0" encoding="UTF-8"?>
<web-app id="starter_anil" 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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Anil-Spring</display-name>
<!-- Servlets -->
<servlet>
<servlet-name>mvc-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mvc-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/classes/applicationContext.xml</param-value>
</context-param>
<!-- Listeners -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
</web-app>
this made exception less start on tomcat but i am little bit confused

If that's where you absolutely must have it, try this:
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="starter_anil" 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 http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
<display-name>Anil-Spring</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</context-param>
<the-rest-of-your-stuff />
</web-app>
You might need a / before applicationContext.xml, but I don't think so.

May be you need to do undeploy. and deploy it. it will work . .
it works for me for the exception
IOException parsing XML document from ServletContext resource [/WEB-INF/classes/spring application config.xml];

WEB-INF/classes/applicationContext.xml for this requirement, you have to put applicationContext.xml in class path i.e.
If you are using maven then put it in src/main/resources folder, otherwise put in the src folder.
Then you will find it inside WEB-INF/classes

Why does spring-ws not detect a spring configuration file if declared in the root context.

I am having a weird problem. I have a my-spring.xml file i would like to add to application but for some reason i can only get it to be detected if it is in the root context.
The following does not detect the my-spring.xml declared in the root context.
<?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/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
version="2.4">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/my-datasources.xml
/WEB-INF/my-spring.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>myservice</servlet-name>
<servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/webservice-config.xml</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>myservice</servlet-name>
<url-pattern>/webservice/*</url-pattern>
</servlet-mapping>
<mime-mapping>
<extension>wsdl</extension>
<mime-type>text/xml</mime-type>
</mime-mapping>
<mime-mapping>
<extension>xsd</extension>
<mime-type>text/xml</mime-type>
</mime-mapping>
</web-app>
If i change it to this then it works (Note the difference in the location of my-spring.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/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
version="2.4">
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/my-datasources.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>myservice</servlet-name>
<servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/webservice-config.xml</param-value>
<param-value>/WEB-INF/my-spring.xml</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>myservice</servlet-name>
<url-pattern>/webservice/*</url-pattern>
</servlet-mapping>
<mime-mapping>
<extension>wsdl</extension>
<mime-type>text/xml</mime-type>
</mime-mapping>
<mime-mapping>
<extension>xsd</extension>
<mime-type>text/xml</mime-type>
</mime-mapping>
</web-app>
I cant really see what i am doing wrong. If i understood the way this works, when declared on the root context (as in the first example) it should be accessed by any resource throught the application. This does not seem to be the case in this example as i cant access the properties defined in the my-spring.xml file from the Controller/Endpoint class.

I think you need commas to separate multiple contextConfigLocation files. Based on this answer: https://stackoverflow.com/a/2725413/116509
Edit: Try removing the preceding / e.g. WEB-INF/my-datasources.xml rather than /WEB-INF/my-datasources.xml

Try something like this
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath:/my-datasources.xml
classpath:/my-spring.xml
</param-value>
</context-param>
But you must have these 2 files in classpath

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