springMVC4集成swagger2

最新推荐文章于 2021-12-30 17:06:22 发布

原创最新推荐文章于 2021-12-30 17:06:22 发布 · 701 阅读

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文章标签：

#spring #swagger

本文介绍了如何将Swagger2集成到SpringMVC4项目中，以生成RESTful API的文档。首先，添加Swagger的相关依赖，然后创建Swagger配置类，设置扫描的控制器包。接着配置web.xml和SpringMVC配置文件，确保DispatcherServlet正确处理静态资源。最后，编写Controller类并访问swagger-ui.html查看生成的API文档。

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本帖子使用的组件版本：
Spring 4.2.1.RELEASE
Swagger 2.4.0

1.Swagger介绍：
Swagger 是一个规范和完整的框架，用于生成、描述、调用和可视化 RESTful 风格的 Web 服务。总体目标是使客户端和文件系统作为服务器以同样的速度来更新。文件的方法，参数和模型紧密集成到服务器端的代码，允许API来始终保持同步。
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2.Swagger如何集成到SpringMVC中：
1）添加maven依赖
  
  <dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger2</artifactId>
<version>2.4.0</version>
  </dependency>
  <dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger-ui</artifactId>
<version>2.4.0</version>
  </dependency>
--springfox-swagger2为Swagger的核心依赖；springfox-swagger-ui为Swagger的UI包，支持通过浏览器（chrome）查看API信息。
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2）增加Swagger配置类
@Configuration
@EnableSwagger2
public class RestApiConfig extends WebMvcConfigurationSupport {
@Bean
public Docket createRestApi1() {
      return new Docket(DocumentationType.SWAGGER_2).groupName("tem1").apiInfo(buildApiInf()).select()
            .apis(RequestHandlerSelectors.basePackage("com.thunisoft.springmvc.business.api.tem1"))//controller路径
            .paths(PathSelectors.any()).build();
}
@Bean
public Docket createRestApi2() {
      return new Docket(DocumentationType.SWAGGER_2).groupName("tem2").apiInfo(buildApiInf()).select()
            .apis(RequestHandlerSelectors.basePackage("com.thunisoft.springmvc.business.api.tem2"))//controller路径
            .paths(PathSelectors.any()).build();
}
private ApiInfo buildApiInf() {
      return new ApiInfoBuilder().title("API").termsOfServiceUrl(" www.xxxx.com ")
            .description("springmvc swagger2").build();
}
}
--这段代码中需要注意的一个地方是groupName，这个相当于分组命名，basePackage为需要扫描的包，每个groupName对应一个basePackage。
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3）配置web.xml文件
<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:/spring/mvc/mvcContext.xml</param-value>
  </init-param>
  <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
  <servlet-name>dispatcherServlet</servlet-name>
  <url-pattern>/</url-pattern>
</servlet-mapping>
--注意配置DispatcherServlet的url-pattern为  /
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4）配置SpringMVC-Context.xml文件：
添加一行配置：<mvc:default-servlet-handler />
--在springMVC-servlet.xml中配置<mvc:default-servlet-handler />后，会在Spring MVC上下文中定义一个org.springframework.web.servlet.resource.DefaultServletHttpRequestHandler，它会像一个检查员，对进入DispatcherServlet的URL进行筛查，如果发现是静态资源的请求，就将该请求转由Web应用服务器默认的Servlet处理，如果不是静态资源的请求，才由DispatcherServlet继续处理。
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5）编写API的Controller类：
在2）步中指定的com.thunisoft.springmvc.business.api.tem1和com.thunisoft.springmvc.business.api.tem2包下面添加所需要的controller类，举例：
@Controller
@RequestMapping(value = "/api1")
public class ApiController1 {
@ApiOperation(value = "获取用户信息", notes = "根据url的id来获取用户详细信息")
@ApiImplicitParam(name = "userId", value = "用户ID", required = true, dataType = "User")
@RequestMapping(value = "/getUser.html", method = RequestMethod.GET)
@ResponseBody
public User getUser(@RequestParam(value = "userId") Integer userId) {
      User u = new User();
      u.setcId("1111");
      u.setcName("test");
      return u;
}
}
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6）访问http://{IP}:{端口}/{应用名称}/swagger-ui.html查看API文档：