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My post request is like this:

POST http://localhost:5000/upload
Content-Type: text/csv
Filename: /path/to/file/filename.csv

I want to receive this file in my code.I have found that the file is received in request.files['file'] in case my file is uploaded using html form where the name associated with the uploaded file is 'file'. But since I am sending the file in the post method of my API, I have no name associated directly to my file object. SO how do I accept the file in my python flask code?

The code I am using is like this but the request.filesobject seems empty after the API hit.

class FileInput(Resource):
    def post(self):
        conn = e.connect()
        if 'file' not in request.files:
            return "No file part."
        file = request.files['file']
        if file.filename == '':
            return "No File Selected"
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return "File Input Successful."
        return "File Input Unsuccessful."
AArias
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Vibhore Kapoor
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  • You mean the method always exits where `return "No File Selected"`? Can you add a `print request.files` right at the beginning of the method and share the output? – AArias Jan 12 '17 at 15:02

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