Could not open ServletContext resource WEB-INF applicationContext xml

Question

Ok  I am 500th user asking this question  I read many answers but still having no luck   parent module pom contains    lt dependency gt       lt groupId gt org springframework lt  groupId gt       lt artifactId gt spring-web lt  artifactId gt       lt version gt   spring framework version  lt  version gt   lt  dependency gt   lt dependency gt       lt groupId gt org springframework lt  groupId gt       lt artifactId gt spring-webmvc lt  artifactId gt       lt version gt   spring framework version  lt  version gt   lt  dependency gt    Child module has maven-jetty-plugin and I run my webapp module with jetty run   web xml defines standard dispatcher module    lt listener gt       lt listener-class gt org springframework web context ContextLoaderListener lt  listener-class gt   lt  listener gt    lt servlet gt       lt servlet-name gt dispatcher lt  servlet-name gt       lt servlet-class gt org springframework web servlet DispatcherServlet lt  servlet-class gt       lt load-on-startup gt 1 lt  load-on-startup gt   lt  servlet gt    lt servlet-mapping gt       lt servlet-name gt dispatcher lt  servlet-name gt       lt url-pattern gt   lt  url-pattern gt   lt  servlet-mapping gt    I have file dispatcher-servlet xml under WEB-INF  though start fails with   FileNotFoundException  Could not open ServletContext resource   WEB-INF applicationContext xml    What is wrong  Documentation and everybody says that Spring MVC will search for XX-servlet xml  where XX is name of servlet  Why does it search for applicationContext xml

User · Accepted Answer

ContextLoaderListener has its own context which is shared by all servlets and filters  By default it will search  WEB-INF applicationContext xml  You can customize this by using   lt context-param gt       lt param-name gt contextConfigLocation lt  param-name gt       lt param-value gt  WEB-INF somewhere-else root-context xml lt  param-value gt   lt  context-param gt    on web xml  or remove this listener if you don t need one

User · Answer

Update  This will create a second context same as in applicationContext xml  or you can add this code snippet to your web xml   lt servlet gt       lt servlet-name gt spring-dispatcher lt  servlet-name gt       lt servlet-class gt org springframework web servlet DispatcherServlet lt  servlet-class gt           lt init-param gt             lt param-name gt contextConfigLocation lt  param-name gt             lt param-value gt classpath applicationContext xml lt  param-value gt           lt  init-param gt       lt load-on-startup gt 1 lt  load-on-startup gt   lt  servlet gt    instead of   lt servlet gt       lt servlet-name gt dispatcher lt  servlet-name gt       lt servlet-class gt org springframework web servlet DispatcherServlet lt  servlet-class gt       lt load-on-startup gt 1 lt  load-on-startup gt   lt  servlet gt

[java] Could not open ServletContext resource [/WEB-INF/applicationContext.xml]

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