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  • swagger2 如何匹配多个controller

    方法一:使用多个controller的共同拥有的父类,即精确到两个controller的上一级

    @Bean
    public Docket createRestApi() {
        return new Docket(DocumentationType.SWAGGER_2)
                .apiInfo(apiInfo())
                .select()
                .apis(RequestHandlerSelectors.basePackage("com.shubing"))
                .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();
    }

    使用以下两种,都是错误的

    @Bean
    public Docket createRestApi() {
        return new Docket(DocumentationType.SWAGGER_2)
                .apiInfo(apiInfo())
                .select()
                .apis(RequestHandlerSelectors.basePackage("com.shubing.*.controller"))
                .paths(PathSelectors.any())
                .build();
    }
    @Bean
    public Docket createRestApi() {
        return new Docket(DocumentationType.SWAGGER_2)
                .apiInfo(apiInfo())
                .select()
                .apis(RequestHandlerSelectors.basePackage("com.shubing.course.controller"))
                .apis(RequestHandlerSelectors.basePackage("com.shubing.user.controller"))
                .paths(PathSelectors.any())
                .build();
    }

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  • 原文地址:https://www.cnblogs.com/acm-bingzi/p/swagger2-controller.html
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