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