I try python Flask and interact with html.
from flask import Flask, render_template, request, redirect, url_for
import os
from werkzeug.utils import secure_filename
# app = Flask(__name__, template_folder="./templates2") # if folders name templates, no need [, template_folder="./templates2"]
app = Flask(__name__)
@app.route('/hello/', methods=['GET', 'POST'])
def hello():
if request.method == 'POST':
name1 = request.form['yourname']
age = request.form['yourage']
return render_template('hello.html', name=name1, age4=age)
return render_template('hello.html')
@app.route('/login/', methods=['GET', 'POST'])
def login():
if request.method == 'POST':
return render_template('login.html')
return render_template('login.html')
''' File Upload '''
# 設置上傳文件夾
UPLOAD_FOLDER = 'uploads'
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
# 確保上傳文件夾存在
os.makedirs(UPLOAD_FOLDER, exist_ok=True)
# 允許上傳的文件類型
ALLOWED_EXTENSIONS = {'txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'}
def allowed_file(filename):
return '.' in filename and \
filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS
@app.route('/upload/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# 檢查是否有文件部分
if 'file' not in request.files:
return 'No file part'
file = request.files['file']
# 如果用戶沒有選擇文件,瀏覽器也會提交一個空的文件名
if file.filename == '':
return 'No selected file'
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('upload_file'))
return render_template('upload.html')
if __name__ == '__main__':
app.run(debug=True)
Also, the sample of ‘Front-End and Back-End Communication.zip’ was tested. Success.
I attended the AWS Bootcamp.
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