python3 flask开发 17 上传文件并显示

# -*- coding: utf-8 -*-
"""
Created on Thu Jan 31 21:42:19 2019
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@author: Administrator
"""
# -*- coding: utf-8 -*-
import os
from flask import Flask, request, url_for, send_from_directory
from werkzeug import secure_filename
#设置上传文件夹地址、允许的文件扩展名、限制文件大小
ALLOWED_EXTENSIONS = set(['png', 'jpg', 'jpeg', 'gif'])

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

#创建一个上传表单
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>
    '''

#检查文件后缀
def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS

#配置一个函数来获取上传文件的url
@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']#当按下提交键后，通过request对象上的files获取文件
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))#使用save()方法保存文件
            file_url = url_for('uploaded_file', filename=filename)
            return html + '<br><img src=' + file_url + '>'
    return html

if __name__ == '__main__':
    app.run()
python3 flask开发 17 上传文件并显示

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