0

我是 Spring MVC 的新手。我newController.java在springproject. 我的代码如下：

@RequestMapping(value = "/Receiver", method = RequestMethod.GET)
public void recvHttpGet(Model model) {
    System.out.println("here get");
    newmethod();
}

@RequestMapping(value = "/Receiver", method = RequestMethod.POST)
public void recvHttpPost(Model model) {
    System.out.println("here post");
    newmethod();
}

@RequestMapping(value = "/", method = RequestMethod.GET)
public String show(Model model) {
    return "index";
}

web.xml

    <?xml version="1.0" encoding="UTF-8"?>
    <web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee"
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
     xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>ClassPath:/spring/applicationContext.xml, ClassPath:/spring/hibernateContext.xml</param-value>
</context-param>

<servlet>
    <servlet-name>appServlet</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>appServlet</servlet-name>
    <url-pattern>/</url-pattern>

</servlet-mapping>

每当我尝试运行它时，index.jsp就会显示页面，但每当我尝试调用/Receiverurl 时，它都会显示 404 错误。请帮我。此外，当我更改recvHttpGet方法 时return "index"，它也会显示 404 错误。也没有向控制台写入任何内容。

我只想检查哪些方法调用所以想在控制台窗口中写入，但它没有显示任何内容。

4

1 回答 1

0

你需要返回一个 JSP 页面，就像return "index";

如果您在视图中有一个receiver.jsp 页面，那么...

@RequestMapping(value = "/Receiver", method = RequestMethod.GET)
public String recvHttpGet(Model model) {

    return "receiver";
}
于 2013-08-24T09:14:56.223 回答