Spring 5：从 url 调用时无法点击控制器

Question

<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
    version="3.1">
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/dispatcher-servlet-context.xml</param-value>
    </context-param>

    <servlet>
        <servlet-name>dispatcher-servlet</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>dispatcher-servlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
    <resource-ref>
        <description>Data source for Analyst</description>
        <res-ref-name>jdbc/ofsaaml</res-ref-name>
        <res-type>javax.sql.DataSource</res-type>
        <res-auth>Container</res-auth>
    </resource-ref>
</web-app>

Answer 1

发布你的完整程序，你可能在你的 url 映射中做错了