相关疑难解决方法(0)

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
Run Code Online (Sandbox Code Playgroud)

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>
Run Code Online (Sandbox Code Playgroud)

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>
Run Code Online (Sandbox Code Playgroud)

HelloWorldController.java

package com.smiechmateusz.controller;

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

import org.springframework.stereotype.Controller;
import …
Run Code Online (Sandbox Code Playgroud)

java spring json

50
推荐指数
7
解决办法
15万
查看次数

HttpMediaTypeNotAcceptableException:在exceptionhandler中找不到可接受的表示形式

我的控制器(Spring 4.1)中有以下图像下载方法:

@RequestMapping(value = "/get/image/{id}/{fileName}", method=RequestMethod.GET)
public @ResponseBody byte[] showImageOnId(@PathVariable("id") String id, @PathVariable("fileName") String fileName) {
    setContentType(fileName); //sets contenttype based on extention of file
    return getImage(id, fileName);
}
Run Code Online (Sandbox Code Playgroud)

以下ControllerAdvice方法应处理不存在的文件并返回json错误响应:

@ExceptionHandler(ResourceNotFoundException.class)
@ResponseStatus(HttpStatus.NOT_FOUND)
public @ResponseBody Map<String, String> handleResourceNotFoundException(ResourceNotFoundException e) {
    Map<String, String> errorMap = new HashMap<String, String>();
    errorMap.put("error", e.getMessage());
    return errorMap;
}
Run Code Online (Sandbox Code Playgroud)

我的JUnit测试完美无瑕

(编辑这是因为扩展.bla:这也适用于appserver):

@Test
public void testResourceNotFound() throws Exception {
    String fileName = "bla.bla";
      mvc.perform(MockMvcRequestBuilders.get("/get/image/bla/" + fileName)
            .with(httpBasic("test", "test")))
            .andDo(print())
            .andExpect(jsonPath("$error").value(Matchers.startsWith("Resource not found")))
            .andExpect(status().is(404));
}
Run Code Online (Sandbox Code Playgroud)

并给出以下输出:

MockHttpServletResponse: …
Run Code Online (Sandbox Code Playgroud)

java junit spring spring-mvc

15
推荐指数
1
解决办法
2万
查看次数

标签 统计

java ×2

spring ×2

json ×1

junit ×1

spring-mvc ×1