java 我们能以某种方式给予Swagger UI文档JSONs吗？

rjjhvcjd 于 12个月前发布在 Java

关注(0)|答案(1)|浏览(145)

我有一个动态创建路由的网关服务。我希望它通过Swagger UI公开其端点。由于我的路由不是静态的（它们随着新服务在Eureka 中的注册和注销而出现和消失），我不能纯粹使用静态yamls（如so）。所以我所做的是：
1.指定配置端点;

springdoc:
  api-docs:
    enabled: false
  swagger-ui:
    enabled: true
    path: /swagger-ui.html
    config-url: /swagger-ui-config

字符串
1.为该终点提供支持;

package by.afinny.apigateway.controller;

import by.afinny.apigateway.model.uiConfig.SwaggerUiConfig;
import by.afinny.apigateway.service.SwaggerUiConfigProvider;
import lombok.RequiredArgsConstructor;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RestController;
import reactor.core.publisher.Mono;

@RestController
@RequiredArgsConstructor
public class SwaggerUiConfigController {
    private final SwaggerUiConfigProvider configProvider;
    @GetMapping("/swagger-ui-config")
    public Mono<SwaggerUiConfig> getConfig() {
        return configProvider.getSwaggerUiConfig();
    }
}

型
1.创建封装配置Swagger UI预期的类（URL名称对的列表）;

package by.afinny.apigateway.model.uiConfig;

import by.afinny.apigateway.model.documentedApplication.SwaggerApplication;
import by.afinny.apigateway.service.SwaggerUiConfigSerializer;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import lombok.Getter;
import lombok.NoArgsConstructor;

import java.util.Collection;

@NoArgsConstructor
@Getter
public class SwaggerUiConfig {
    @JsonProperty("urls")
    @JsonSerialize(contentUsing = SwaggerUiConfigSerializer.class)
    private Collection<SwaggerApplication> swaggerApplications;

    public SwaggerUiConfig(Collection<SwaggerApplication> swaggerApplications) {
        this.swaggerApplications = swaggerApplications;
    }

    public static SwaggerUiConfig from(Collection<SwaggerApplication> swaggerApplications) {
        return new SwaggerUiConfig(swaggerApplications);
    }
}

package by.afinny.apigateway.service;

import by.afinny.apigateway.model.documentedApplication.SwaggerApplication;
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.SerializerProvider;
import lombok.SneakyThrows;

import java.io.IOException;
import java.text.MessageFormat;

public class SwaggerUiConfigSerializer extends JsonSerializer<SwaggerApplication> {
    @Override
    @SneakyThrows
    public void serialize(SwaggerApplication swaggerApplication, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException {
        jsonGenerator.writeStartObject();
        jsonGenerator.writeStringField("url", MessageFormat.format("/{0}{1}", swaggerApplication.getName(), SwaggerApplication.SWAGGER_DOC_PATH));
        jsonGenerator.writeStringField("name", swaggerApplication.getName());
        jsonGenerator.writeEndObject();
    }
}

的字符串
1.提供对应于这些url的路由;路由只是转发到lb://SERVICE-NAME/v3/api-docs

// I'll omit that code

型
我不会提供所有的代码，因为这并不重要，我也不想让你不知所措。
但问题是我已经获取了那些JSON文档来构建我的业务路由（每个端点的路由），它们已经存储在内存中（并进行了更新）。
我能以某种方式直接给予Swagger UI文档，而不是告诉它在哪里找到它们吗？

Java

来源：https://stackoverflow.com/questions/77637384/can-we-somehow-give-swagger-ui-documentation-jsons-directly

1条答案

按热度按时间

ogq8wdun1#

我能找到的最直接的方法是为配置和文档本身编写端点。

package com.example.dynamicgateway.controller;

import com.example.dynamicgateway.model.uiConfig.SwaggerUiConfig;
import com.example.dynamicgateway.service.swaggerUiSupport.SwaggerUiSupport;
import lombok.RequiredArgsConstructor;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RestController;
import reactor.core.publisher.Mono;

@RestController
@RequiredArgsConstructor
public class SwaggerUiConfigController {
    private final SwaggerUiSupport uiSupport;

    @GetMapping("/swagger-ui-config")
    public Mono<SwaggerUiConfig> getConfig() {
        return uiSupport.getSwaggerUiConfig();
    }
}

package com.example.dynamicgateway.controller;

import com.example.dynamicgateway.service.swaggerUiSupport.SwaggerUiSupport;
import io.swagger.v3.oas.models.OpenAPI;
import lombok.RequiredArgsConstructor;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PathVariable;
import org.springframework.web.bind.annotation.RestController;
import reactor.core.publisher.Mono;

@RestController
@RequiredArgsConstructor
public class SwaggerDocController {
    private final SwaggerUiSupport uiSupport;

    @GetMapping("{application-name}/doc")
    public Mono<OpenAPI> getSwaggerAppDoc(@PathVariable("application-name") String applicationName) {
        // return cached OpenAPI object right away
        return uiSupport.getSwaggerAppDoc(applicationName);
    }
}

package com.example.dynamicgateway.model.uiConfig;

import com.example.dynamicgateway.model.documentedApplication.SwaggerApplication;
import com.example.dynamicgateway.service.swaggerUiSupport.SwaggerUiConfigSerializer;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import lombok.Getter;
import lombok.NoArgsConstructor;

import java.util.Collection;

@NoArgsConstructor
@Getter
public class SwaggerUiConfig {
    @JsonProperty("urls")
    @JsonSerialize(contentUsing = SwaggerUiConfigSerializer.class)
    private Collection<SwaggerApplication> swaggerApplications;

    public SwaggerUiConfig(Collection<SwaggerApplication> swaggerApplications) {
        this.swaggerApplications = swaggerApplications;
    }

    public static SwaggerUiConfig from(Collection<SwaggerApplication> swaggerApplications) {
        return new SwaggerUiConfig(swaggerApplications);
    }
}

package com.example.dynamicgateway.service.swaggerUiSupport;

import com.example.dynamicgateway.model.documentedApplication.SwaggerApplication;
import com.fasterxml.jackson.core.JsonGenerator;
import com.fasterxml.jackson.databind.JsonSerializer;
import com.fasterxml.jackson.databind.SerializerProvider;
import lombok.SneakyThrows;

import java.text.MessageFormat;

public class SwaggerUiConfigSerializer extends JsonSerializer<SwaggerApplication> {
    @Override
    @SneakyThrows
    public void serialize(SwaggerApplication swaggerApplication, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) {
        jsonGenerator.writeStartObject();
        jsonGenerator.writeStringField("url", MessageFormat.format("/{0}/doc", swaggerApplication.getName()));
        jsonGenerator.writeStringField("name", swaggerApplication.getName());
        jsonGenerator.writeEndObject();
    }
}

springdoc:
  swagger-ui:
    enabled: true
    path: /swagger-ui.html
    config-url: /swagger-ui-config

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java 我们能以某种方式给予Swagger UI文档JSONs吗？

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