从 openapi swagger 上传的文件未在后端 python 函数中收到

Question

我正在尝试使用 Openapi 3.0 和 swagger UI 从用户那里获取文件。但是我没有在我的 python 函数中获取该文件进行处理。下面是我的代码：

code.py

def get_file():
    try:
        file=request.files.getlist('file')[0]
        with open(file, 'r') as fp:
            files = {"file": (file, fp)}
            response = requests.post(server, files=files)
            return response.json()
    except Exception as exc:
        return exc

api.yaml

  /get-result:
    post:
      summary: "A function to get file"
      operationId: "code.get_file"
      requestBody:
      content:
        application/json:
          schema:
            type: string
              format: binary
      responses:
        200:
          description: "executed successfully"
          content:
            application/json:
              schema:
                $ref: "#/components/schemas/myschema"
        500:
          description: Server is down.

我已经提到了这个link：Upload a file in Swagger and receive at Flask backend 然而，这是针对 Openapi 2.0 的，并没有帮助，因为我使用的是 openapi 3.0

Answer 1

检查是否

request.files['file']

可以从请求中获取文件，我不确定行

file=request.files.getlist('file')[0]

实际上会得到正确的文件（或只是一个列表？）

Answer 2

下面的代码更改帮助我解决了这个问题：

api.yaml

/get-result:
    post:
      summary: "A function to get file"
      operationId: "code.get_file"
      requestBody:
        content:
          multipart/form-data:
            schema:
              type: object
              properties:
                file:
                  type: string
                  format: binary
      responses:
        200:
          description: "executed successfully"
          content:
            application/json:
              schema:
                $ref: "#/components/schemas/myschema"
        500:
          description: Server is down.

下面code.py必须将参数'file'传递给函数，这也是openapi规范中的字段名称。使用 post 请求

发送文件指针和文件名也很重要

code.py

def get_file(file):
    try:
        fp=file.read()
        file.save(file.filename)
        response = requests.post(server, files={"file":(file.filename,fp)})
        return response.json()
    except Exception as exc:
        return exc

从 openapi swagger 上传的文件未在后端 python 函数中收到

file uploaded from openapi swagger not getting received at backend python function

python

rest

flask

swagger

openapi