2013-02-25 32 views
10

我有一個從JQuery.post()調用的彈簧控制器。當它被調用時,控制器的方法被調用並返回。但是,在後臺,Spring會更改URL並調用服務器增益。服務器響應與404彈簧控制器404在POST方法調用後重新調用

我認爲這是爲了應對春節試圖找到一個查看POST方法已被處理後。

如何阻止Spring控制器執行此操作。

這裏是我的春天控制器:

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); 
    } 
} 

這裏是我的應用程序上下文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> 

這裏是我的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> 

這裏是我的從我的HTML文件調用JQuery:

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功能只是做一個alert("DONE")

所以當JQuery的方法調用的網址是:

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

,並在服務器端的System.out.println(...)方法打印出預期的數據。

但是在Firebug,服務器發回一個404

於是我打開與Spring記錄,並得到這樣的:

[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 

爲響應URL POST請求(/ jQuery的/數據/人/放)時,正確的方法是發現和調用(線1〜7),但然後在第8行,這會改變網址/jquery/data/person/person/put向前春天到InternalResourceView,這不能被發現。

如何阻止Spring嘗試查找視圖。我希望它做的就是乾乾淨淨地完成工作。

感謝您的幫助。

回答

6

我相信,如果你有一個空或返回void類型,Spring將盡力解決基於請求URL的看法。我認爲這裏的適當形式是簡單地返回一個OK頁面,因爲這看起來不是JSON或類似的東西。或者,只需使用@ResponseBody標記它並返回一個空字符串即可。

+1

感謝您的回覆。答案是添加@ResponseBody。春天似乎處理無效返回罰款。 – 2013-02-26 10:25:15

+2

@AdamDavies將其作爲答案發布並將其標記爲接受 – 2013-02-27 12:50:57

11

已解決。

的問題是作爲#CodeChimp建議,但我仍然希望返回類型爲void。

我加入了@ResponseBodyaddPerson方法,一切都工作得很好:

@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); 
} 

線索從http://static.springsource.org/spring/docs/3.2.x/spring-framework-reference/html/mvc.html#mvc-ann-responsebody來了。雖然文檔不清楚無效返回會發生什麼。剛剛嘗試過,它工作。

+0

我猜測void或null返回將與空字符串相同。 Spring將簡單地在響應主體中沒有任何東西。我發現春天的人很聰明,並且傾向於在他們的解決方案中思考所有可能的角度。 – CodeChimp 2013-02-27 14:05:38

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