Java 招摇过市文档:对响应内容的描述

Java 招摇过市文档:对响应内容的描述,java,jersey,jax-rs,swagger,Java,Jersey,Jax Rs,Swagger,我有一个简单的fileUpload rest API,它将上传的文件存储在数据库中,并返回数据库中上传文件的unque引用(uuid) 我使用swagger生成API文档,它的工作方式很有魅力,但我找不到添加描述文本来描述响应内容的方法 这是我其他人的签名: @POST @Path("/upload") @Consumes(ExtendedMediaType.MULTIPART_FORM_DATA) @ApiOperation( value = "Save an image",

我有一个简单的fileUpload rest API,它将上传的文件存储在数据库中,并返回数据库中上传文件的unque引用(uuid)

我使用swagger生成API文档,它的工作方式很有魅力,但我找不到添加描述文本来描述响应内容的方法

这是我其他人的签名:

@POST
@Path("/upload")
@Consumes(ExtendedMediaType.MULTIPART_FORM_DATA)
@ApiOperation(
        value = "Save an image",
        notes = "Save the uploaded image to database.",
        response = String.class)
@ApiResponses(value = {
        @ApiResponse(code = 200, message = "The uploaded image has been saved.", response = String.class),
        @ApiResponse(code = 500, message = "Error while trying to resize the image.<p>Internal error code: 2103", response = a.b.ErrorInfo.class)})
@Override
public String uploadImage(
        @ApiParam(value = "file to upload", required = true) @FormDataParam("file") final InputStream inputStream,
        @ApiParam(value = "details of the uploaded file", required = true) @FormDataParam("file") final FormDataContentDisposition fileDetail) {

    //return UUID.randomUUID().toString();
}
我已经查看了文档,但没有找到任何与此主题相关的内容


我可以将此信息放在
ApiResponse
HTTP 200旁边,但我不确定此解决方案是否正确。

根据我使用swagger的经验,我认为必须在ApiResponse的notes字段中指定此信息,如下所示

@ApiOperation(
        value = "Save an image",
        notes = "Returns the unique id of the uploaded image",
        response = String.class)
@ApiOperation(
        value = "Save an image",
        notes = "Returns the unique id of the uploaded image",
        response = String.class)