如何为 swagger REST API 文档生成 java 客户端代码
我的场景如下.
我有一个招摇的 .json 例如:http://petstore.swagger.io/v2/swagger.json我想为上面的 REST API 使用生成的 java 客户端，例如:
I have a swagger .json eg.: http://petstore.swagger.io/v2/swagger.json I want to use a generated java client for the REST API above, like:
PetApi petApi = new PetApi();
Pet pet = new Pet;
pet.setName("cica");
pet.setId(1L);
petApi.addPet(pet);
System.out.println(petApi.getById(1L));`
扩展输出:cica，新宠物按照REST API实现存储.
Expexted output: cica and the new pet is stored according to the REST API implmentation.
我已使用以下命令成功为 petstore 生成了服务器存根:
I have successfully generated server stub for the petstore with the command:
java -jar modules/swagger-codegen-cli/target/swagger-codegen-cli.jar generate
-i http://petstore.swagger.io/v2/swagger.json
-l spring-mvc
-o samples/server/petstore/spring-mvc
但是这个 maven 项目代码是一个服务器代码.它在 PetApi.java 中有类似 @RequestMapping 的注解，并且还有一个 WebMvcConfiguration.class.
But this maven project code is a server code. It has annotations like @RequestMapping in PetApi.java and also has a WebMvcConfiguration.class.
我不想拥有服务器存根.我想要一个用于 petstore REST API 的客户端库.
I do not want to have a server-stub. I want to have a client-library for the petstore REST API.
有没有工具可以为我生成合适的客户端库?我应该修改服务器存根，因此它具有所有模型还是应该使用简单的 springRestTemplate?
Is there a tool which can generate the appropriate client library for me? Should I modify the server-stub, hence it has all the models or should I use a simple springRestTemplate?
感谢您的回答！
推荐答案
我觉得你对Swagger Codegen的参数-l的取值不对(你用的是spring-mvc 是一种服务器端技术).您可以尝试使用值 java.
I think that you don't use the right value for the parameter -l of Swagger Codegen (you use spring-mvc which is a server-side technology). You could try to use the value java.
您还可能注意到有一个工具，Restlet Studio，它允许从 Swagger 生成代码内容.对于 Java，它主要依赖于 Restlet 框架，但我认为它可以满足您的需求.
You could also notice that there is a tool, the Restlet Studio, that allows to generate code from Swagger content. For Java, it mainly relies on the Restlet framework but I think that it could suit your needs.
希望对你有帮助蒂埃里
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