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

Question

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.

I think you should add file: before value. file:/WEB-INF/spring-context.xml — Harshal Khachane
– Harshal Khachane, Commented Mar 14, 2020 at 5:07
@HarshalKhachane It's gives me the previous error with this one now: _GRAVE: StandardWrapper.Throwable org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from URL [file:/WEB-INF/spring-context.xml]; nested exception is java.io.FileNotFoundException: \WEB-INF\spring-context.xml _ — JamesB
– JamesB, Commented Mar 14, 2020 at 6:03

Govinda Sakhare · Accepted Answer · 2020-03-14 06:34:37Z

1

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>

edited Mar 14, 2020 at 6:34

answered Mar 14, 2020 at 6:01

Govinda Sakhare

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3 Comments

JamesB Over a year ago

The error keeps on....

ERROR ContextLoader:319 - Context initialization failed 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]

JamesB Over a year ago

No just spring-context.xml and web.xml.

Govinda Sakhare Over a year ago

Would it be possible to share the codebase? May be on GitHub.

Marco Behler · Accepted Answer · 2020-03-14 09:17:50Z

0

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>

answered Mar 14, 2020 at 9:17

Marco Behler

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2 Comments

JamesB Over a year ago

param-value is null.

Marco Behler Over a year ago

Give it a try :)

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