Question

有人可以帮我这个吗？

My Controller类看起来像这样，我创建了客户模型类..

/**
 * Handles requests for the application home page.
 */
@Controller
@RequestMapping("/customer")
public class CustomerController {

    @RequestMapping(method = RequestMethod.GET)
    public ModelAndView student() {
        return new ModelAndView("customer", "command", new Customer());
    }

    @RequestMapping(value = "/addCustomer", method = RequestMethod.POST)
       public String addStudent(@ModelAttribute Customer customer, 
       ModelMap model) {

        model.addAttribute("customerName", customer.getCustomerName());
        model.addAttribute("emailId", customer.getEmailId());
        model.addAttribute("sex", customer.getSex());

        return "customerDetails";
    }
}

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" 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_2_5.xsd">

    <display-name>Customer Form Handling</display-name>


    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/root-context.xml</param-value>
    </context-param>

    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- Processes application requests -->
    <servlet>
        <servlet-name>Customer</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/Customer/servlet-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>Customer</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

</web-app>

JSP页面

customer.jsp

<%@taglib uri="http://www.springframework.org/tags/form" prefix="form"%>
<html>
<head>
    <title>Customer Form Handling</title>
</head>
<body>

<h2>Customer Information</h2>
<form:form method="POST" commandName = "command" action="/addCustomer">
   <table>
    <tr>
        <td><form:label path="customerName">customerName</form:label></td>
        <td><form:input path="customerName" /></td>
    </tr>
    <tr>
        <td><form:label path="emailId">emailId</form:label></td>
        <td><form:input path="emailId" /></td>
    </tr>
    <tr>
        <td><form:label path="sex">sex</form:label></td>
        <td><form:input path="sex" /></td>
    </tr>
    <tr>
        <td colspan="2">
            <input type="submit" value="Submit"/>
        </td>
    </tr>
</table>  
</form:form>
</body>
</html>

customerDetails.jsp

<%@taglib uri="http://www.springframework.org/tags/form" prefix="form"%>
<html>
<head>
    <title>Customer Form Handling</title>
</head>
<body>

<h2>Customer Detail Information</h2>
   <table>
    <tr>
        <td>CustomerName</td>
        <td>${customerName}</td>
    </tr>
    <tr>
        <td>emailId</td>
        <td>${emailId}</td>
    </tr>
    <tr>
        <td>sex</td>
        <td>${sex}</td>
    </tr>
</table>  
</body>
</html>

Servlet的context.xml中

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->
    <annotation-driven />

    <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <!-- <resources mapping="/resources/**" location="/resources/" /> -->

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/" />
        <beans:property name="suffix" value=".jsp" />
    </beans:bean>

    <context:component-scan base-package="com.customerinfo.controller" />

</beans:beans>

但是当我在Tomcat Server中运行此应用程序时..第一个url指向 localhost:8080/controller/。

如果我追加localhost:8080/controller/customer，我会收到第一个表单页面。

但是一旦我点击提交..它说找不到页面错误。

Answer 1

这是一个相对路径问题。您的表单操作为/addCustomer（具有/前缀），如果您解决它，则为http://localhost:8080/addCustomer。您想要的可能是http://localhost:8080/appname/customer/addCustomer。

在某些情况下，只需将其更改为customer/addCustomer即可解决此问题，但如果您的页面也可以通过http://localhost:8080/appname/customer/访问（请注意尾随斜杠），那么这可能是一个问题。相对路径将转换为http://localhost:8080/appname/customer/customer/addCustomer

当然，现在你可以考虑只做/appname/customer/addCustomer并解决问题，但事实上你现在正在硬编码上下文路径名。如果有一天上下文路径发生了变化，那么所有这些代码都会中断。

我喜欢使用一种方法，因此我的JSP可以通过定义根变量

来确定上下文路径

<c:set var="root" value="${pageContext.request.contextPath}"/>
...
<form:form action="${root}/customer/addCustomer">

Answer 2

尝试

<form:form method="POST" commandName = "command" action="addCustomer">

而不是

<form:form method="POST" commandName = "command" action="/addCustomer">

Spring MVC，表单提交不起作用

2 个答案: