Java Swagger返回xml格式而不是json。我需要json格式来查看
在我的项目中,我使用的是swagger。这对我来说很好。我使用的是swagger版本0.5.0。我需要生成输出的json格式,但swagger只生成xml格式。如何重新处理此错误。我使用的是spring和rest服务Java Swagger返回xml格式而不是json。我需要json格式来查看,java,json,rest,swagger,Java,Json,Rest,Swagger,在我的项目中,我使用的是swagger。这对我来说很好。我使用的是swagger版本0.5.0。我需要生成输出的json格式,但swagger只生成xml格式。如何重新处理此错误。我使用的是spring和rest服务 pom.xml <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:sche
pom.xml
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>csrfspring</groupId>
<artifactId>csrfspring</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<build>
<sourceDirectory>src</sourceDirectory>
<plugins>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.1</version>
<configuration>
<source>1.7</source>
<target>1.7</target>
</configuration>
</plugin>
<plugin>
<artifactId>maven-war-plugin</artifactId>
<version>2.4</version>
<configuration>
<warSourceDirectory>WebContent</warSourceDirectory>
<failOnMissingWebXml>false</failOnMissingWebXml>
</configuration>
</plugin>
</plugins>
</build>
<properties>
<jdk.version>1.6</jdk.version>
<spring.version>4.2.1.RELEASE</spring.version>
<jstl.version>1.2</jstl.version>
<javax.servlet.version>3.1.0</javax.servlet.version>
</properties>
<dependencies>
<dependency>
<groupId>com.mangofactory</groupId>
<artifactId>swagger-springmvc</artifactId>
<version>0.5.0</version>
</dependency>
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-core</artifactId>
<version>2.0.2</version>
</dependency>
<dependency>
<groupId>javax.ws.rs</groupId>
<artifactId>javax.ws.rs-api</artifactId>
<version>2.0.1</version>
</dependency>
<dependency>
<groupId>com.wordnik</groupId>
<artifactId>swagger-core</artifactId>
<version>1.0</version>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>javax.servlet-api</artifactId>
<version>${javax.servlet.version}</version>
<scope>provider</scope>
</dependency>
<!-- Spring 3 dependencies -->
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-core</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-web</artifactId>
<version>${spring.version}</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>${spring.version}</version>
</dependency>
<!-- jstl for jsp page -->
<dependency>
<groupId>jstl</groupId>
<artifactId>jstl</artifactId>
<version>${jstl.version}</version>
</dependency>
<dependency>
<groupId>javax</groupId>
<artifactId>javaee-web-api</artifactId>
<version>7.0</version> <!-- Put here the version of your Java EE app, in my case 7.0 -->
<scope>provided</scope>
</dependency>
</dependencies>
</project>
Spring xml
<?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:p="http://www.springframework.org/schema/p"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.2.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.2.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd">
<mvc:default-servlet-handler />
<mvc:annotation-driven />
<context:component-scan base-package="com.examle.anand" />
<bean class="com.mangofactory.swagger.configuration.DocumentationConfig" />
<context:property-placeholder location="classpath:/swagger.properties" />
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/jsp/" />
<property name="suffix" value=".jsp" />
</bean>
</beans>
controller in java
package com.examle.anand;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.ui.ModelMap;
import com.wordnik.swagger.annotations.Api;
@Controller
@Api(value = "hello", description = "it simple test api")
@RequestMapping("/hello")
public class HelloController{
@RequestMapping(value="/test",method = RequestMethod.GET)
public String printHello(ModelMap model) {
System.out.println("TEsted i am inside controller");
model.addAttribute("message", "Hello Spring MVC Framework!");
return "hello";
}
}
pom.xml
4.0.0
csrfspring
csrfspring
0.0.1-快照
战争
src
maven编译器插件
3.1
1.7
1.7
maven战争插件
2.4
网络内容
错误的
1.6
4.2.1.1发布
1.2
3.1.0
芒果工厂
大摇大摆
0.5.0
com.fasterxml.jackson.core
杰克逊核心
2.0.2
javax.ws.rs
javax.ws.rs-api
2.0.1
com.wordnik
大摇大摆的核心
1
javax.servlet
javax.servlet-api
${javax.servlet.version}
供应商
org.springframework
弹簧芯
${spring.version}
org.springframework
弹簧网
${spring.version}
org.springframework
SpringWebMVC
${spring.version}
jstl
jstl
${jstl.version}
爪哇
javaeewebapi
7
假如
SpringXML
java中的控制器
包com.examle.anand;
导入org.springframework.stereotype.Controller;
导入org.springframework.web.bind.annotation.RequestMapping;
导入org.springframework.web.bind.annotation.RequestMethod;
导入org.springframework.ui.ModelMap;
导入com.wordnik.swagger.annotations.Api;
@控制器
@Api(value=“hello”,description=“it简单测试Api”)
@请求映射(“/hello”)
公共类Hello控制器{
@RequestMapping(value=“/test”,method=RequestMethod.GET)
公共字符串printHello(ModelMap模型){
System.out.println(“已测试,我在控制器内部”);
addAttribute(“message”,“HelloSpringMVC框架!”);
回复“你好”;
}
}
只需通过公共XML库从XML创建Java对象,然后通过Jackson库创建JSON即可。查看并阅读斯威格规范。但是如果你想得到完整的JSON支持,那么就使用RAML。这是一大堆代码。你把范围缩小到问题发生的地方了吗?我建议通过.pom.xml来帮助我们,我使用的依赖关系是什么。spring xml如何配置我的java控制器的组件扫描。控制器定义了我使用的控制器。上面的代码非常简单