在POST方法调用后,Spring Controller 404重新调整

时间:2013-02-25 12:58:34

标签: jquery spring-mvc

我有一个从JQuery.post()调用的Spring控制器。调用它时,调用控制器的方法并返回。但是,在后台,Spring会更改URL并调用服务器增益。服务器以404响应。

我认为这是对Spring在POST方法处理后尝试查找View的回应。

如何阻止Spring控制器执行此操作。

这是我的Spring Controller:

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.PathVariable;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;

import java.util.ArrayList;
import java.util.List;

@Controller
@RequestMapping("/person")
public class DataController {

  private List<Person> people = new ArrayList<Person>();

  @RequestMapping(value="put", method = RequestMethod.POST)
  public void addPerson(@ModelAttribute("person") Person person){
    System.out.println(">>>>>>> person: " + person);
    System.out.println(">>>>>>>>> " + person.getFirstName());
    people.add(person);
  }
}

这是我的应用程序上下文xml文件:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       xmlns:context="http://www.springframework.org/schema/context"
       xsi:schemaLocation="
         http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
         http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
         http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">

    <context:component-scan base-package="uk.co.jeeni" />

    <mvc:annotation-driven />

</beans>

这是我的web.xml文件:

<?xml version="1.0" encoding="ISO-8859-1" ?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
                http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
         version="2.4">

    <servlet>
        <servlet-name>dispatcherServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath*:applicationContext-web.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>dispatcherServlet</servlet-name>
        <url-pattern>/data/*</url-pattern>
    </servlet-mapping>
</web-app>

这是我的HTML文件中的JQuery调用:

function attachSendDataEvent(){
    $("#sendData").click(function(){

        var data = "firstName=" + $("#firstName").val() + "&" +
                "lastName=" + $("#lastName").val() + "&" +
                "address=" + $("#address").val() + "&" +
                "postcode=" + $("#postcode").val();

        $.post("data/person/put",
                data,
                dataSentOK
        );
    });

    return false;
}

dataSentOK函数只执行alert("DONE")

因此,当调用的URL的JQuery方法是:

http://localhost:8080/jquery/data/person/put

在服务器端,System.out.println(...)方法按预期打印数据。

但是在Firebug中,服务器会发回404。

所以我打开了使用Spring的日志记录并得到了这个:

[01] DispatcherServlet [DEBUG] DispatcherServlet with name 'dispatcherServlet' processing POST request for [/jquery/data/person/put]
[02] AbstractHandlerMethodMapping [DEBUG] Looking up handler method for path /person/put
[03] AbstractHandlerMethodMapping [DEBUG] Returning handler method [public void uk.co.jeeni.DataController.addPerson(uk.co.jeeni.Person)]
[04] AbstractBeanFactory [DEBUG] Returning cached instance of singleton bean 'dataController'
[05] DispatcherServlet [DEBUG] Rendering view [org.springframework.web.servlet.view.InternalResourceView: name 'person/put'; URL [person/put]] in DispatcherServlet with name 'dispatcherServlet'
[06] AbstractView [DEBUG] Added model object 'org.springframework.validation.BindingResult.person' of type [org.springframework.validation.BeanPropertyBindingResult] to request in view with name 'person/put'
[07] AbstractView [DEBUG] Added model object 'person' of type [uk.co.jeeni.Person] to request in view with name 'person/put'
[08] InternalResourceView [DEBUG] Forwarding to resource [person/put] in InternalResourceView 'person/put'
[09] DispatcherServlet [DEBUG] DispatcherServlet with name 'dispatcherServlet' processing POST request for [/jquery/data/person/person/put]
[10] AbstractHandlerMethodMapping [DEBUG] Looking up handler method for path /person/person/put
[11] AbstractHandlerMethodMapping [DEBUG] Did not find handler method for [/person/person/put]
[12] DispatcherServlet [ WARN] No mapping found for HTTP request with URI [/jquery/data/person/person/put] in DispatcherServlet with name 'dispatcherServlet'
[13] FrameworkServlet [DEBUG] Successfully completed request
[14] FrameworkServlet [DEBUG] Successfully completed request

响应URL POST请求(/ jquery / data / person / put),找到并调用了正确的方法(第1行到第7行),然后Spring在第8行转发到InternalResourceView,这会将网址更改为/jquery/data/person/person/put,但无法找到。

如何阻止Spring尝试查找视图。我想要它做的就是干净利落地完成。

感谢您的帮助。

2 个答案:

答案 0 :(得分:13)

解决。

问题是#CodeChimp建议的,除了我还想要一个void返回类型。

我将@ResponseBody添加到addPerson方法,一切正常:

@RequestMapping(value="put", method = RequestMethod.POST)
**@ResponseBody**
public void addPerson(@ModelAttribute("person") Person person){
  System.out.println(">>>>>>> person: " + person);
  System.out.println(">>>>>>>>> " + person.getFirstName());
  people.add(person);
}

线索来自http://static.springsource.org/spring/docs/3.2.x/spring-framework-reference/html/mvc.html#mvc-ann-responsebody。虽然文档不清楚void返回会发生什么。刚试了一下就行了。

答案 1 :(得分:6)

我相信如果你有一个null或void返回类型,Spring会尝试根据请求URL解析视图。我认为这里适当的形式是简单地返回一个OK页面,因为这似乎不是JSON或类似的东西。或者,只需使用@ResponseBody标记它并返回一个空字符串。