前言:项目中有多个模块,所以有多个controller层。我是用方案二的
正文:
方案一:使用多个controller的共同拥有的父类
@Bean public Docket createRestApi() { return new Docket(DocumentationType.SWAGGER_2) .apiInfo(apiInfo()) .select() .apis(RequestHandlerSelectors.basePackage("com.xx")) .paths(PathSelectors.any()) .build(); }
方法二:指定所有controller的都实现的一个接口,比如@RestController
@Bean public Docket createRestApi() { return new Docket(DocumentationType.SWAGGER_2) .apiInfo(apiInfo()) .select() .apis(RequestHandlerSelectors.withClassAnnotation(RestController.class)) .paths(PathSelectors.any()) .build(); }
错误的两种写法
@Bean public Docket createRestApi() { return new Docket(DocumentationType.SWAGGER_2) .apiInfo(apiInfo()) .select() .apis(RequestHandlerSelectors.basePackage("com.xx.*.controller")) .paths(PathSelectors.any()) .build(); }
@Bean public Docket createRestApi() { return new Docket(DocumentationType.SWAGGER_2) .apiInfo(apiInfo()) .select() .apis(RequestHandlerSelectors.basePackage("com.xx.course.controller")) .apis(RequestHandlerSelectors.basePackage("com.xx.user.controller")) .paths(PathSelectors.any()) .build(); }
参考博客:
swagger2 如何匹配多个controller - 贾树丙 - 博客园
http://www.cnblogs.com/acm-bingzi/p/swagger2-controller.html