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调查 Springfox 和 Swagger UI，但我面临一个问题。我使用 Spring Boot REST 示例项目作为我的 PoC 的基础。我正在运行 JDK 8，并且该项目利用了 Gradle。

首先，这里是项目的文件内容：

构建.gradle

buildscript {
    repositories {
        mavenCentral()
    }
    dependencies {
        classpath("org.springframework.boot:spring-boot-gradle-plugin:1.2.7.RELEASE")
    }
}

apply plugin: 'java'
apply plugin: 'eclipse'
apply plugin: 'idea'
apply plugin: 'spring-boot'

jar {
    baseName = 'gs-rest-service'
    version =  '0.1.0'
}

repositories {
    mavenCentral()
    jcenter()
}

sourceCompatibility = 1.8
targetCompatibility = 1.8

dependencies {
    compile("org.springframework.boot:spring-boot-starter-web")
    compile("io.springfox:springfox-swagger2:2.2.2")
    compile("io.springfox:springfox-swagger-ui:2.2.2")
    testCompile("junit:junit")
}

task wrapper(type: Wrapper) {
    gradleVersion = '2.3'
}

GreetingController.java

package hello;

import java.util.concurrent.atomic.AtomicLong;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class GreetingController {

    private static final String template = "Hello, %s!";
    private final AtomicLong counter = new AtomicLong();

    @RequestMapping("/greeting")
    public Greeting greeting(@RequestParam(value="name", defaultValue="World") String name) {
        return new Greeting(counter.incrementAndGet(),
                            String.format(template, name));
    }
}

问候语.java

package hello;

public class Greeting {

    private final long id;
    private final String content;

    public Greeting(long id, String content) {
        this.id = id;
        this.content = content;
    }

    public long getId() {
        return id;
    }

    public String getContent() {
        return content;
    }
}

应用程序.java

package hello;

import static com.google.common.collect.Lists.newArrayList;
import static springfox.documentation.schema.AlternateTypeRules.newRule;

import java.time.LocalDate;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.context.annotation.Bean;
import org.springframework.http.ResponseEntity;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.context.request.async.DeferredResult;

import springfox.documentation.builders.PathSelectors;
import springfox.documentation.builders.RequestHandlerSelectors;
import springfox.documentation.builders.ResponseMessageBuilder;
import springfox.documentation.schema.ModelRef;
import springfox.documentation.schema.WildcardType;
import springfox.documentation.spi.DocumentationType;
import springfox.documentation.spring.web.plugins.Docket;
import springfox.documentation.swagger2.annotations.EnableSwagger2;

import com.fasterxml.classmate.TypeResolver;

@SpringBootApplication
@EnableSwagger2
public class Application {

    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }

    @Autowired
    private TypeResolver typeResolver;

    @Bean
    public Docket greetingApi() {
        return new Docket(DocumentationType.SPRING_WEB)
            .select()
            .apis(RequestHandlerSelectors.any())
            .paths(PathSelectors.any())
            .build()
            .pathMapping("/")
            .directModelSubstitute(LocalDate.class, String.class)
            .genericModelSubstitutes(ResponseEntity.class)
            .alternateTypeRules(newRule(typeResolver.resolve(DeferredResult.class,
                typeResolver.resolve(ResponseEntity.class, WildcardType.class)),
                typeResolver.resolve(WildcardType.class)))
            .useDefaultResponseMessages(false)
            .globalResponseMessage(RequestMethod.GET,
                newArrayList(new ResponseMessageBuilder()
                    .code(500)
                    .message("500 message")
                    .responseModel(new ModelRef("Error"))
                    .build()))
            .enableUrlTemplating(true);
    }

}

这是我面临的问题。当我构建并运行应用程序时，我可以成功导航到 Swagger UI 页面 ( http://localhost:8080/swagger-ui.html )。当我展开问候控制器时，我看到了不同的方法并展开“get /greeting{?name}”。Get 部分有以下内容：

Response Class (Status 200)

Model

{
  "content": "string",
  "id": 0
}

Response Content Type: */*

Parameters
parameter = name, value = World, parameter type = query, data type = string

当我单击“试用”按钮时，我看到以下内容：

curl = curl -X GET --header "Accept: */*" "http://localhost:8080/greeting{?name}?name=World"

request url = http://localhost:8080/greeting{?name}?name=World

repsonse body = {
  "timestamp": 1446418006199,
  "status": 404,
  "error": "Not Found",
  "message": "No message available",
  "path": "/greeting%7B"
}

response code = 404

response headers = {
  "server": "Apache-Coyote/1.1",
  "content-type": "application/json;charset=UTF-8",
  "transfer-encoding": "chunked",
  "date": "Sun, 01 Nov 2015 22:46:46 GMT"
}

乍一看，出于某种原因，Springfox/Swagger 似乎没有正确替换 {?name} 的占位符。我的问题是，如果这确实是问题，我该如何配置它，以便我可以从 Swagger UI 页面成功测试服务？

4

1 回答 1

5

在您的Application班级中将其更改enableUrlTemplating为 false 将解决您的问题。

@Bean
public Docket greetingApi() {
    return new Docket(DocumentationType.SPRING_WEB)
        //...
        .enableUrlTemplating(false);
}

只是那面旗帜的一点背景。该标志是为了支持RFC 6570，如果没有它，仅通过查询字符串参数不同的操作将不会按照规范正确显示。在 swagger 规范的下一次迭代中，有计划解决这个问题。这就是enableUrlTemplating被标记为孵化功能的原因。

于 2015-11-01T23:09:57.033 回答