錯誤的鏈接位置錯誤

Question

我有一個web應用程序。我的控制器看起來是這樣的：

@Controller 
@RequestMapping(value = "/") 
public class MainController {  

    @RequestMapping(method = RequestMethod.GET) 
    public String index() { 
     return "index"; 
    } 

    @RequestMapping(value = "/add", method = RequestMethod.POST) 
    public String add(ModelMap model, User user) { 
     userService.create(user); 
     model.addAttribute("message", "User successfully created."); 
     return "index"; 
    } 
} 

而且我有這樣一種形式：

<form:form method="POST" action="/add" commandName="user"> 

的web.xml

<servlet> 
    <servlet-name>client</servlet-name> 
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> 
    <load-on-startup>1</load-on-startup> 
</servlet> 

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

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

<listener> 
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> 
</listener> 

我的應用程序通過鏈接訪問：http://localhost:8080/Client
但是，當我提交表格提交至http://localhost:8080/add應該是http://localhost:8080/Client/add
我想，我可以在每個鏈接之前使用${pageContext.request.contextPath}，但有沒有其他方法？

Answer 1

試試這個：

<form:form method="POST" action="${requestContext.pathToServlet}/add" commandName="user"> 

瞭解更多：

• https://jira.springsource.org/browse/SPR-8863
• https://jira.springsource.org/browse/SPR-8684

相關的問題：

• Spring MVC appending context path to form action
• Spring MVC: Url path appending when posting the form

Answer 2

使用c:url可將上下文根自動添加到您的URL。

<c:url var="formUrl" value="/add" /> 
<form:form method="POST" action="${formUrl}" commandName="user">