从 openapi 上传的文件 swagger 未在后端收到 python function

Question

I am trying to take a file from user using Openapi 3.0 and swagger UI. However i am not getting that file for processing in my python function. Below is my code:

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.

I have already referred this link: Upload a file in Swagger and receive at Flask backend However this is for Openapi 2.0 and didnt help as I am using openapi 3.0

Answer 1

check if

request.files['file']

can get the file from the request, I'm not sure if the line

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

will actually get the correct file (or only a list?)

Answer 2

Below code changes helped me to get this resolved:

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.

In below code.py it was essential to pass the parameter 'file' to the function which is also the field name in openapi specification. Also its important to send file pointer and filename with the post request

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