how to return the path of file when uploading in python flask code example

Example 1: create internal etl for people to upload data with python and flask

from flask import Flask, render_template, requestfrom werkzeug import secure_filenameapp = Flask(__name__)@app.route('/upload')def upload_file():   return render_template('upload.html')	@app.route('/uploader', methods = ['GET', 'POST'])def upload_file():   if request.method == 'POST':      f = request.files['file']      f.save(secure_filename(f.filename))      return 'file uploaded successfully'		if __name__ == '__main__':   app.run(debug = True)

Example 2: flask upload

def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS

@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        # check if the post request has the file part
        if 'file' not in request.files:
            flash('No file part')
            return redirect(request.url)
        file = request.files['file']
        # if user does not select file, browser also
        # submit an empty part without filename
        if file.filename == '':
            flash('No selected file')
            return redirect(request.url)
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file',
                                    filename=filename))
    return '''
    <!doctype html>
    <title>Upload new File</title>
    <h1>Upload new File</h1>
    <form method=post enctype=multipart/form-data>
      <input type=file name=file>
      <input type=submit value=Upload>
    </form>
    '''

Tags:

Python Example

Recent Posts