java为什么大摇大摆的ui不显示带注释的REST方法？

1 周，2 日 Questions & Answers 15346

我无法配置swagger以查看我的REST方法。我在Eclipse和Tomcat7中工作。我有以下简单的REST方法/类：

package com.rest;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import com.wordnik.swagger.annotations.Api;
import com.wordnik.swagger.annotations.ApiOperation;

@Api( value="/", description="Say hello class")
@Path("/")
public class Hello {

    @GET
    @Path("/hello")
    @ApiOperation(value="/hello", notes="hello method")
    public String sayHello() {
        return "Hello World!";
    }
}

我正在使用以下网站。xml

   <?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:javaee="http://java.sun.com/xml/ns/javaee"
    xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0">
    <display-name>SwaggerTest</display-name>
    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
    <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>com.wordnik.swagger.jaxrs.json,com.rest</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.provider.classnames</param-name>
            <param-value>
                com.wordnik.swagger.jersey.listing.ApiListingResourceJSON,
                com.wordnik.swagger.jersey.listing.JerseyApiDeclarationProvider,
                com.wordnik.swagger.jersey.listing.JerseyResourceListingProvider
            </param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/api/*</url-pattern>
    </servlet-mapping>


    <servlet>
        <servlet-name>SwaggerJerseyJaxrsConfig</servlet-name>
        <servlet-class>com.wordnik.swagger.jersey.config.JerseyJaxrsConfig</servlet-class>
        <init-param>
            <param-name>api.version</param-name>
            <param-value>0.0.1</param-value>
        </init-param>
        <init-param>
            <param-name>swagger.api.basepath</param-name>
            <param-value>http://localhost:8080/api/</param-value>
        </init-param>
        <load-on-startup>2</load-on-startup>
    </servlet>

</web-app>

rest服务在http://localhost:8080/SwaggerTest/api/sayHello上可用，并在浏览器中显示正确的消息。该服务的招摇过市规范可在http://localhost:8080/SwaggerTest/api-docs上找到。然而，所有返回的都是

{"apiVersion":"0.0.1","swaggerVersion":"1.2","apis":[{"path":"/","description":"Say hello class"}]}

GET sayHello（）方法发生了什么？或者这就是它应该返回的全部

任何帮助都将不胜感激。提前谢谢

大卫

另外，maven依赖项是

<dependencies>
    <dependency>
        <groupId>org.glassfish.jersey.core</groupId>
        <artifactId>jersey-server</artifactId>
        <version>2.7</version>
    </dependency>

    <dependency>
        <groupId>org.glassfish.jersey.media</groupId>
        <artifactId>jersey-media-multipart</artifactId>
        <version>2.7</version>
    </dependency>

    <dependency>
        <groupId>org.glassfish.jersey.media</groupId>
        <artifactId>jersey-media-json-jackson</artifactId>
        <version>2.7</version>
    </dependency>

    <dependency>
        <groupId>com.wordnik</groupId>
        <artifactId>swagger-jersey2-jaxrs_2.10</artifactId>
        <version>1.3.12</version>
    </dependency>

    <dependency>
        <groupId>org.webjars</groupId>
        <artifactId>swagger-ui</artifactId>
        <version>2.1.1</version>
    </dependency>

    <dependency>
        <groupId>com.wordnik</groupId>
        <artifactId>swagger-servlet_2.10</artifactId>
        <version>1.3.12</version>
    </dependency>

    <dependency>
        <groupId>org.glassfish.jersey.containers</groupId>
        <artifactId>jersey-container-servlet-core</artifactId>
        <version>2.7</version>
    </dependency>


</dependencies>

<?xml version="1.0" encoding="UTF-8"?> <web-app xmlns:javaee="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" version="3.0"> <display-name>RTSServices</display-name> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <servlet> <servlet-name>Jersey REST Service</servlet-name> <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> <init-param> <param-name>jersey.config.server.provider.packages</param-name> <param-value>com.ibm.dipo.rest, com.wordnik.swagger.jaxrs.listing</param-value> </init-param> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>Jersey REST Service</servlet-name> <url-pattern>/api/BlueHound/*</url-pattern> </servlet-mapping> <servlet> <servlet-name>SwaggerJerseyJaxrsConfig</servlet-name> <servlet-class>com.wordnik.swagger.jersey.config.JerseyJaxrsConfig</servlet-class> <init-param> <param-name>api.version</param-name> <param-value>0.0.1</param-value> </init-param> <init-param> <param-name>swagger.api.basepath</param-name> <param-value>http://localhost:8080/RTSServices/api/BlueHound/</param-value> </init-param> <load-on-startup>2</load-on-startup> </servlet> </web-app>

共 (2) 个答案

# 1 楼答案

我没有大摇大摆地告诉泽西检查正确的包裹。尤其是com。沃德尼克。大摇大摆贾克斯。必须添加以下列表：

<init-param>
    <param-name>jersey.config.server.provider.packages</param-name>
    <param-value>com.ibm.dipo.rest,
                com.wordnik.swagger.jaxrs.listing</param-value>
</init-param>

然后我

将swagger dist树（https://github.com/swagger-api/swagger-ui/tree/master/dist）加载到我的WebContent目录中
将其重命名为api文档

修改索引。html指向swagger的api文档位置：

$（函数（）{ var url=window。地方搜索匹配（/url=（[^&；]+）/）

  if (url && url.length > 1) {
    url = decodeURIComponent(url[1]);
  } else {
    url = "http://petstore.swagger.io/v2/swagger.json";
    url = "../../api/BlueHound/api-docs"
  }

有了这些更改，我的招摇文档可以在http://localhost:8080/RTSServices/api-docs/index.html

完整的网络。xml如下所示：

# 2 楼答案

问题是您使用的是swagger core 1.3，它生成了swagger 1.2定义。Swagger 1.2不太喜欢基于root（/）的API。如果将@Api的value更改为“/”以外的任何内容，仍然可以显示它。这不会影响API本身，只会影响文档的托管方式

例如，如果您给它“/root”的value，然后转到http://localhost:8080/SwaggerTest/api-docs/root，您将看到您公开的服务

此外，您正在使用旧版本的swagger core和swagger general。看起来您正在尝试与Jersey集成，因此您可以遵循https://github.com/swagger-api/swagger-core/wiki/Swagger-Core-Jersey-1.X-Project-Setup-1.5作为您的集成指南。这就产生了Swagger 2.0，它与根资源没有相同的问题

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java为什么大摇大摆的ui不显示带注释的REST方法？

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# 2 楼答案