如何在Flask图片上传API中接收客户端传递的文件名并以此保存图片

Let's tweak your code so the server saves the uploaded image using the filename the client sends over. Here's the step-by-step fix for both your server and client scripts:

Modified Server Code (server.py) #

We'll enable the commented-out logic to receive the filename parameter and use it when saving the image:

from flask import Flask
from flask_restful import Resource, Api, reqparse
import werkzeug

app = Flask(__name__)
api = Api(app)

class UploadImage(Resource):
    def post(self):
        parser = reqparse.RequestParser()
        # Add argument to accept the filename from the client
        parser.add_argument('FNAME', required=True)
        parser.add_argument('file', type=werkzeug.datastructures.FileStorage, location='files')
        args = parser.parse_args()
        imageFile = args['file']
        # Fetch the client-specified filename
        filename = args['FNAME']
        print(f"Received filename to save: {filename}")
        # Save the image using the provided filename instead of hardcoding
        imageFile.save(filename)
        return {"status": "success", "saved_as": filename}

api.add_resource(UploadImage, '/uploadimage')

if __name__ == '__main__':
    from waitress import serve
    print("Running....")
    serve(app,host="0.0.0.0",port=8080)
    print("Stopped....")

Key Updates: #

• Uncommented parser.add_argument('FNAME', required=True) to capture the filename parameter from the client
• Retrieved the client's desired filename with filename = args['FNAME']
• Changed imageFile.save('test.jpg') to imageFile.save(filename) to use the custom name
• Added a JSON response to confirm the image was saved with the correct name (helpful for debugging)

Modified Client Code (client.py) #

We'll add a data parameter to the POST request to send your desired filename to the server:

import requests

# Open your local image file
dfile = open("test.jpg", "rb")
url = "http://127.0.0.1:8080/uploadimage"

# Set the filename you want the server to use for the saved image
desired_filename = "my_custom_photo.jpg"

# Send both the image file and the filename parameter
test_res = requests.post(url, files={"file": dfile}, data={"FNAME": desired_filename})
print(test_res.json())  # Print the server's confirmation response

if test_res.ok:
    print(" File uploaded successfully ! ")
else:
    print(" Please Upload again ! ")

Key Updates: #

• Added a data parameter to requests.post() that passes the FNAME key with your custom filename
• Updated the print statement to show the server's JSON response (so you can verify the saved filename)
• If you want to use the original name of your local file, replace desired_filename with dfile.name

How It Works: #

1. The client sends both the image file and the desired filename in a single POST request
2. The server parses both pieces of data
3. The server saves the image using the exact filename provided by the client

内容的提问来源于stack exchange，提问作者user19728274