java Swagger 示例帖子正文 - 如何显示 JSON 正文 - Swagger-注释
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Swagger example post body - how to show JSON Body- Swagger-annotations
提问by Sum
Requirement: I have a POST method which takes the input JSON as a String and passes it to another microservice. I don't want to create an Object (Bean) of this input JSON.
要求：我有一个 POST 方法，它将输入的 JSON 作为字符串并将其传递给另一个微服务。我不想创建此输入 JSON 的对象（Bean）。
method:
方法：
@ApiOperation(notes = "example" value = "/example", consumes = ".." , method= "..")
@RequestMapping(name = "xxx" value ="/hello" ..)
@ApiResponses(..)
public @ResponseBody String getXXX (@Apiparam(name="JSONrequest", required = true) @RequestBody String JSONrequest){
}
Problem: The generated Swagger doesn't show the input as a JSON model where all the JSON attributes are displayed.
问题：生成的 Swagger 没有将输入显示为显示所有 JSON 属性的 JSON 模型。
Expectation: I want to display my Swagger Something like this :
期望：我想显示我的 Swagger 像这样的东西：
Definately I am missing the key thing. Any thoughts?
毫无疑问，我错过了关键的事情。有什么想法吗？
回答by Andrei Sfat
If changing from Stringto a concrete object is not okay (although that's what I would recommend you to do since it's cleaner), you can try using @ApiImplicitParams(check out their documentation)
如果从更改String为具体对象是不行的（尽管我建议您这样做，因为它更干净），您可以尝试使用@ApiImplicitParams（查看他们的文档）
@ApiOperation(notes = "example" value = "/example", consumes = ".." , method= "..")
@ApiImplicitParams({
@ApiImplicitParam(name = "Object", value = "Object to be created", required = true, dataType = "your.package.BodyClass", paramType = "body")
})
@RequestMapping(name = "xxx" value ="/hello" ..)
@ApiResponses(..)
public @ResponseBody String getXXX (@Apiparam(name="JSONrequest", required = true) @RequestBody String JSONrequest){
}
(not sure if you still need the @Apiparam(name="JSONrequest", required = true)bit from the method parameter)
（不确定您是否仍然需要@Apiparam(name="JSONrequest", required = true)方法参数中的位）
