相关疑难解决方法(0)

Spring JSON请求获得406(不可接受)

这是我的javascript:

    function getWeather() {
        $.getJSON('getTemperature/' + $('.data option:selected').val(), null, function(data) {
            alert('Success');                               
        });
    }
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这是我的控制器:

@RequestMapping(value="/getTemperature/{id}", headers="Accept=*/*", method = RequestMethod.GET)
@ResponseBody
public Weather getTemparature(@PathVariable("id") Integer id){
    Weather weather = weatherService.getCurrentWeather(id);
        return weather;
}
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为spring-servlet.xml

<context:annotation-config />
<tx:annotation-driven />
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得到此错误:

GET http://localhost:8080/web/getTemperature/2 406 (Not Acceptable)
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头:

响应标题

Server  Apache-Coyote/1.1
Content-Type    text/html;charset=utf-8
Content-Length  1070
Date    Sun, 18 Sep 2011 17:00:35 GMT
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请求标题

Host    localhost:8080
User-Agent  Mozilla/5.0 (Windows NT 6.1; WOW64; rv:6.0.2) Gecko/20100101 Firefox/6.0.2
Accept  application/json, text/javascript, */*; q=0.01
Accept-Language en-us,en;q=0.5
Accept-Encoding gzip, …
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javascript java ajax spring json

83
推荐指数
7
解决办法
18万
查看次数

Spring MVC + JSON = 406不可接受

我正在尝试生成一个简单的JSON响应.现在我得到406 Not Acceptable错误.Tomcat说:"此请求标识的资源只能根据请求"接受"标题生成具有不可接受特征的响应." 即使我的Accept标题是

Accept:text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
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在tomcat/lib中,我有所有Tomcat jar,Spring jar和jackson-all-1.9.0.jar.我正在使用Spring 3.2.2和Tomcat 7.

我知道这个问题已经讨论了很多次,但没有一个解决方案适合我.

web.xml中

<web-app id="WebApp_ID" version="2.4" 
    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">

  <display-name>Spring Web MVC Application</display-name>

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

  <servlet-mapping>
    <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
  </servlet-mapping>

</web-app>
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调度员servlet.xml中

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:mvc="http://www.springframework.org/schema/mvc" 
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
     http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context-3.0.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">

    <bean id="viewResolver"
        class="org.springframework.web.servlet.view.InternalResourceViewResolver" >
        <property name="prefix">
            <value>/WEB-INF/pages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
    </bean>
 <context:component-scan base-package="com.smiechmateusz.controller" />
 <context:annotation-config />

    <mvc:annotation-driven />

</beans>
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HelloWorldController.java

package com.smiechmateusz.controller;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.springframework.stereotype.Controller;
import …
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java spring json

50
推荐指数
7
解决办法
15万
查看次数

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