Swagger-UI(Docker) - “试一试”时不保持基本路径

问题描述 投票:0回答:1

我有一个JAX-RS应用程序,它使用Swagger2导出openapi.json:

http://localhost:8080/notification/rest/openapi.json

并公开OpenAPI:

{
  "openapi" : "3.0.1",
  "info" : {
    "title" : "Notification Module",
    "version" : "1.0"
  },
  "paths" : {
    "/sms/send" : {
      "post" : {
        "operationId" : "send_1",
        "requestBody" : {
          "content" : {
            "application/json" : {
              "schema" : {
                "$ref" : "#/components/schemas/Sms"
              }
            }
          }
        },
        "responses" : {
          "default" : {
            "description" : "default response",
            "content" : {
              "application/json" : {
                "schema" : {
                  "$ref" : "#/components/schemas/SendSmsResult"
                }
              }
            }
          }
        }
      }
    }

    ...
  }
}

现在我试图在Docker容器中显示Swagger-UI: docker run -p 80:8080 -e URL="http://localhost:8080/notification/rest/openapi.json" swaggerapi/swagger-ui

Swagger2正常打开。但它试图要求的“试一试”功能: http://localhost:8080/sms/send

缺少上下文和基本路径/notification/rest/

是否可以配置swagger-ui来创建URL请求: http://localhost:8080/notification/rest/sms/send

docker jax-rs swagger swagger-ui
1个回答
0
投票

解决了。

只需将servers配置为openapi-configuration.yaml即可。

resourcePackages:
  - com.mycompany.app.rest
prettyPrint: true
cacheTTL: 0
buildContext: true
openAPI:
  servers:
    - url: http://localhost:8080/notification
  info:
    version: '1.0'
    title: Notification Module

另外,使用@Context ServletConfig参数添加构造函数非常重要:

@ApplicationPath("/rest")
public class App extends Application {
    public App(@Context ServletConfig servletConfig) {
    }
}
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