IT_TI
2016-05-29 17:37
<?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"> <!-- 配置DispatcherServlet --> <servlet> <!-- 配置springMVC需要加载的配置文件 spring-dao.xml,spring-service.xml,spring-web.xml Mybatis Spring SpringMVC --> <servlet-name>seckill</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <init-param> <param-name>contextConfigLoaction</param-name> <param-value>classpath:spring/spring-*.xml</param-value> </init-param> </servlet> <servlet-mapping> <servlet-name>seckill</servlet-name> <!-- 默认匹配所有的请求 --> <url-pattern>/</url-pattern> </servlet-mapping> </web-app>
这个是web.xml中的配置
但是程序运行是报异常
org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from ServletContext resource [/WEB-INF/seckill-servlet.xml]; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/WEB-INF/seckill-servlet.xml]
请问老师这是什么原因
<param-name>contextConfigLoaction</param-name>单词拼写错误location,仔细一点
Java高并发秒杀API之web层
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