如何为swagger注释Play 2 webapp模型？
如何为swagger注释Play 2 webapp模型？
提问于 2013-02-24 19:53:22
我是一名android/java开发人员，刚接触Play2框架。我正在尝试用swagger为我的RESTful应用编程接口生成文档。
我已经设法将swagger包含到我的Play2 api应用程序中，并生成简单的api-docs.json。我唯一缺少的部分是模型描述。我在/controllers和/models中有相应的用户控制器和用户模型。
@Api(value = "/user", listingPath = "/api-docs.{format}/user", description = "User registration and authorisation")
public class User extends Controller {
@POST
@ApiOperation(value = "Create user", notes = "Used to register new user.")
@ApiParamsImplicit(@ApiParamImplicit(name = "body", value = "Created user object", required = true, dataType = "User", paramType = "body"))
@BodyParser.Of(BodyParser.Json.class)
public static Result createUser() {
JsonNode json = request().body().asJson();
ObjectNode result = Json.newObject();
JsonNode body = json.findPath("body");
if(body.isMissingNode()) {
result.put("status", "KO");
result.put("message", "Missing parameter [body]");
return badRequest(result);
}
JsonNode name = body.get("name");
if(name == null) {
result.put("status", "KO");
result.put("message", "Missing parameter [body.name]");
return badRequest(result);
}
result.put("status", "OK");
result.put("message", "Hello " + name.getTextValue());
return ok(result);
}
}我已经尝试过完全像在example中一样注释模型
@XmlRootElement(name = "User")
public class User {
public String name;
@XmlElement(name = "name")
public String getName() {
return name;
}
}结果是：
{
apiVersion: "beta",
swaggerVersion: "1.1",
basePath: "http://localhost:9000",
resourcePath: "/user",
apis: [
{
path: "/user",
description: "User registration and authorisation",
operations: [
{
httpMethod: "POST",
summary: "Create user",
notes: "Used to register new user.",
responseClass: "void",
nickname: "createUser",
parameters: [
{
name: "body",
description: "Created user object",
paramType: "body",
required: true,
allowMultiple: false,
dataType: "User"
}
]
}
]
}
]
}有什么想法吗？
回答 1
Stack Overflow用户
回答已采纳
发布于 2013-02-24 22:24:32
我自己找到了答案。当模型被用作返回值时，swagger似乎会确认该模型，例如responseClass
@ApiOperation( value = "Find quiz by ID",
notes = "Returns a quiz with given ID",
responseClass = "models.Quiz" )
@ApiErrors( value = {
@ApiError(code = 400, reason = "Invalid ID supplied"),
@ApiError(code = 404, reason = "Quiz not found") })
public static Result getQuizById(
@ApiParam(value = "ID of question that needs to be fetched", required = true) @PathParam("quizId")
String quizId) {
ObjectNode result = Json.newObject();
return ok(result);
}只需添加这样的方法，api-docs.json中就会出现相应的模型。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/15051456
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