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[英]How to resolve BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/dispatcher-servlet.xml]?
[英]IOException parsing XML document from ServletContext resource [/WEB-INF/spring-dispatcher-servlet.xml]
從控制台收到此錯誤:
*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]*
這是我收到的錯誤:
<?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>
我在做
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-context.xml</param-value>
</init-param>
從 springmvc 更改默認上下文,但它不起作用。 已經在這里接受了一些建議,將 servlet-name 標記寫入文件 name-context.xml 約定,同樣的錯誤。
您應該嘗試更改 servlet 名稱。 預期的 Spring web-context 元數據 XML 文件名應該是servletname-servlet.xml
,這是直到 Spring 4 的預期文件名,不確定它是否在 spring 5 中更改。。 由於您的 servlet 名稱是springmvc
,文件名應該是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>
使您的 web.xml 看起來像這樣,直接取自 Spring 文檔:
<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>
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