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如何将python flask代码连接到html代码_Python_Flask_Pythonanywhere - Fatal编程技术网

如何将python flask代码连接到html代码

如何将python flask代码连接到html代码,python,flask,pythonanywhere,Python,Flask,Pythonanywhere,我正在尝试让用户上传照片,然后在提交时将这些照片放入pythonawhere的文件夹中。我意识到,即使网站正在运行,我的html和python代码似乎没有连接,因此,当我上传pdf时,即使这不是允许的扩展,我的webapp也会接收它,但它不会处理python代码上的条件 import os from flask import Flask, flash, request, redirect, url_for, render_template, send_from_directory from we

我正在尝试让用户上传照片,然后在提交时将这些照片放入pythonawhere的文件夹中。我意识到,即使网站正在运行,我的html和python代码似乎没有连接,因此,当我上传pdf时,即使这不是允许的扩展,我的webapp也会接收它,但它不会处理python代码上的条件

import os
from flask import Flask, flash, request, redirect, url_for, render_template, send_from_directory
from werkzeug.utils import secure_filename

UPLOAD_FOLDER = '/home/HassanSherien/mysite/Shape_Image'
ALLOWED_EXTENSIONS = {'png', 'jpg', 'jpeg'}

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
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':
        # 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 a 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)
            flash(file)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file',filename=filename))


    elif request.method == "GET":
        return render_template("overview_main_page.html", comments=Comment.query.all())
    return

@app.route('/uploads/<filename>') #, #methods=["GET", "POST"])
def uploaded_file(filename):
    return send_from_directory(app.config['UPLOAD_FOLDER'],
                               filename)
导入操作系统
从flask导入flask、flash、请求、重定向、url、呈现模板、从目录发送
从werkzeug.utils导入安全文件名
UPLOAD_FOLDER='/home/HassanSherien/mysite/Shape_Image'
允许的_扩展={'png','jpg','jpeg'}
app=烧瓶(名称)
app.config['UPLOAD\u FOLDER']=UPLOAD\u FOLDER
允许的def_文件(文件名):
在文件名和\
filename.rsplit('.',1)[1].lower()在允许的扩展名中
@app.route('/upload',methods=['GET','POST'])
def upload_文件():
如果request.method==“POST”:
#检查post请求是否包含文件部分
如果“文件”不在request.files中:
闪存(“无文件部分”)
返回重定向(request.url)
file=request.files['file']
#如果用户未选择文件,则浏览器也会
#提交一个没有文件名的空部分
如果file.filename=='':
闪存('没有选定的文件')
返回重定向(request.url)
如果文件和允许的文件(file.filename):
filename=secure\u文件名(file.filename)
闪存(文件)
保存(os.path.join(app.config['UPLOAD\u FOLDER'],文件名))
返回重定向(url_for('uploaded_file',filename=filename))
elif request.method==“GET”:
返回render\u模板(“overview\u main\u page.html”,comments=Comment.query.all())
返回
@app.route('/uploads/')#,#方法=[“GET”,“POST”])
def上传文件(文件名):
从_目录返回发送_(app.config['UPLOAD_FOLDER'],
文件名)


HTML表单上的操作没有发布到您显示的视图中——它的操作设置为路由“/”,而不是“/upload”。我想我已经解决了,谢谢@吉尔斯托马斯
  <div class="w3-panel">
    <div class="w3-row-padding" style="margin:0 -16px">
      <div class="w3-third">
       <form action="http://hassansherien.pythonanywhere.com" method="post">
           <input type="file" name="file" value="fileupload" id="file_upload">
           <input type='submit' />