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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?

IOException parsing XML document from ServletContext resource [/WEB-INF/spring-dispatcher-servlet.xml]

Receiving this error from console:
*org.springframework.beans.factory.BeanDefinitionStoreException:
IOException parsing XML document from ServletContext resource
[/WEB- INF/spring-dispatcher-servlet.xml]; nested exception is
java.io.FileNotFoundException: Could not open ServletContext resource
[/WEB-INF/spring-dispatcher-servlet.xml]*
That's the error I receive from:
<?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>fj21-tarefas</display-name>
<welcome-file-list>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
I doing
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-context.xml</param-value>
</init-param>
to change the default context from springmvc, but it is not working. Already took some advice here to write a servlet-name tag to the file name-context.xml convention, same error.

You should try changing the servlet name. Expected Spring web-context metadata XML file name should servletname-servlet.xml, this was expected filename till Spring 4, not sure if it is changed in spring 5.
. As your servlet name is springmvc, filename should be springmvc-servlet.xml.
<servlet>
<servlet-name>springmvc</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/springmvc-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springmvc</servlet-name>
<url-pattern>/app-name/*</url-pattern>
</servlet-mapping>

Make your web.xml look like this, taken straight from the Spring documentation:
<web-app>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-context.xml</param-value>
</context-param>
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value></param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>

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 integrate swagger with tomcat?

I am new to swagger, I am trying to swagger to my tomcat application, and below 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>Restful Web Application</display-name>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:xxx/xxx/xxx/xml/applicationContext.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>jersey-helloworld-serlvet</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>xxx.xxx.xxx.resource</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet>
<servlet-name>JerseyJaxrsConfig</servlet-name>
<servlet-class>com.wordnik.swagger.jersey.config.JerseyJaxrsConfig</servlet-class>
<init-param>
<param-name>api.version</param-name>
<param-value>1.0.0</param-value>
</init-param>
<init-param>
<param-name>swagger.api.basepath</param-name>
<param-value>http://localhost:8088/sms</param-value>
</init-param>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-helloworld-serlvet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
And I added Swagger Ui to my WebContent/WEB-INF folder, and added
swagger-annotations_2.9.1
swagger-jersey-jaxrs_2.10
swagger-core_2.10 jars to my class path
I am getting the below error:
java.lang.NoClassDefFoundError: com/wordnik/swagger/config/ConfigFactory
Appreciate your help on this.

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