问使用swagger-php将嵌套属性序列化为form-data
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Stack Overflow用户

提问于 2019-01-29 07:56:08

回答 1查看 2.9K关注 0票数 1

这就是在OpenApi https://github.com/OAI/OpenAPI-Specification/blob/master/versions/3.0.0.md中编码对象示例的方式

requestBody:
  content:
    multipart/mixed:
      schema:
        type: object
        properties:
          id:
            # default is text/plain
            type: string
            format: uuid
          address:
            # default is application/json
            type: object
            properties: {}
          historyMetadata:
            # need to declare XML format!
            description: metadata in XML format
            type: object
            properties: {}
          profileImage:
            # default is application/octet-stream, need to declare an image type only!
            type: string
            format: binary
      encoding:
        historyMetadata:
          # require XML Content-Type in utf-8 encoding
          contentType: application/xml; charset=utf-8
        profileImage:
          # only accept png/jpeg
          contentType: image/png, image/jpeg
          headers:
            X-Rate-Limit-Limit:
              description: The number of allowed requests in the current period
              schema:
                type: integer

我正在尝试实现相同的功能，但使用swagger-php。我不知道的是如何在@OA\MediaType中传递encodings来将test属性编码为multipart/form-data，因为默认情况下编码为application/json

例如：

 * @OA\Post(
 *     path="/admin/test",
 *     summary="Create new Test",
 *     description="Will attempt to create a new Test",
 *     tags={"Admin Test"},
 *     @OA\RequestBody(
 *          @OA\MediaType(
 *              mediaType="multipart/form-data",
 *              encoding={}
 *              @OA\Schema(
 *                  type="object",
 *                  @OA\Property(
 *                      property="test",
 *                      type="object",
 *                      description="test"
 *                      ref="#/components/schemas/MyTestSchema"
 *                  )
 *              )
 *      )

他们在这里有一些例子：

https://github.com/zircote/swagger-php/tree/master/Examples

但是我没有找到任何关于编码的例子

在这里，字段被定义为https://github.com/zircote/swagger-php/blob/master/src/Annotations/MediaType.php

   /**
     * A map between a property name and its encoding information.
     * The key, being the property name, must exist in the schema as a property.
     * The encoding object shall only apply to requestBody objects when the media type is multipart or application/x-www-form-urlencoded.
     */
    public $encoding = UNDEFINED;

我试过encoding={"recommended"={"contentType"="multipart/form-data"}}，但它没有用。

php

swagger

openapi

swagger-php

swagger-3.0

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回答 1

Stack Overflow用户

回答已采纳

发布于 2019-02-12 16:50:06

我认为唯一的解决方案是将所有字段：

 * @OA\Post(
 *     path="/admin/test",
 *     summary="Create new Test",
 *     description="Will attempt to create a new Test",
 *     tags={"Admin Test"},
 *     @OA\RequestBody(
 *          @OA\MediaType(
 *              mediaType="multipart/form-data",
 *              @OA\Schema(
 *                  @OA\Property(
 *                      property="test[name]",
 *                      type="string",
 *                      description="name"
 *                  ),
 *                  @OA\Property(
 *                      property="test[desc]",
 *                      type="string",
 *                      description="description"
 *                  )
 *              )
 *          )
 *      )