网页小白求助一个基本的 flask 网页框架

V2EX = way to explore

V2EX 是一个关于分享和探索的地方

现在注册

已注册用户请登录

› Flask Framework

这是一个创建于 2626 天前的主题，其中的信息可能已经有所发展或是发生改变。

我需要做一个简单的网页

功能和这个网页类似, http://zhcn.109876543210.com/

选择文件---> 后端处理 ---> 处理完毕 ---> 提供下载链接。

对界面完全没有要求, 只想求一个具有类似功能的代码的框架, 自己学习了 flask

以下是我的代码, 现在不知道如何在处理完毕之后提供下载的链接。

多谢各位

# -*- coding: utf-8 -*-
import os
from flask import Flask, request, url_for, send_from_directory
from werkzeug import secure_filename

ALLOWED_EXTENSIONS = set(['png', 'jpg', 'jpeg', 'gif','txt'])

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = os.getcwd()
app.config['MAX_CONTENT_LENGTH'] = 16 * 1024 * 1024


html = '''
    <!DOCTYPE html>
    <title>Upload File</title>
    <h1>图片上传</h1>
    <form method=post enctype=multipart/form-data>
         <input type=file name=file>
         <input type=submit value=上传>
    </form>
    <form method=post enctype=multipart/form-data>
         <input type=file name=file>
         <input type=submit value=上传>
    </form>
    '''

# html = html + '<a href=r"\\DESKTOP-0EVA06J\\Origami\\test.txt" download>下载</a>'
with open('template.html', encoding = 'utf-8') as the_file:
   html = the_file.read()


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


@app.route('/uploads/<filename>')
def uploaded_file(filename):
    return send_from_directory(app.config['UPLOAD_FOLDER'],
                               filename)


@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        file = request.files['file']
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            print('the file name will be ' + os.path.join(app.config['UPLOAD_FOLDER'], filename))
#the file will be here.

            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            file_url = url_for('uploaded_file', filename=filename)
            # print('the file_url will be ' + file_url)
            return html + '<br><img src=' + file_url + '>'
    return html


if __name__ == '__main__':
    # app.run(host='192.168.0.176', port=5001)
    app.run(extra_files = html)
    # app.run()

2 条回复 • 2017-09-22 17:42:39 +08:00

pipapa

2017-09-22 17:12:33 +08:00

前端利用 ajax,POST 数据,获取返回值

gosky

2017-09-22 17:42:39 +08:00

对 flask 了解有限
post /upload
处理，保存文件到{uuid}.data
get /results/{uuid}.data