from flask import Flask, request, render_template,redirect, url_for
from werkzeug.utils import secure_filename
import os
from flask import send_from_directory # send_from_directory可以从目录加载文件app = Flask(__name__)#UPLOAD_FOLDER = 'media' # 注意：要提前在根目录下新建media文件，否则会报错
'''
报错路径找不到：
要解决此问题，可以将反斜杠替换为正斜杠，或者使用原始字符串字面值指定文件路径.
通过在字符串前添加r，它就变成了一个原始字符串文字，并且反斜杠将被解释为文字反斜杠而不是转义字符。
'''
#UPLOAD_FOLDER = 'D:\pythonXangmu\pythonproject01\flask\caiquan2\media'
UPLOAD_FOLDER = r'D:\pythonXangmu\pythonproject01\flask\caiquan2\media'ALLOWED_EXTENSIONS = set(['txt', 'pdf', 'png', 'jpg', 'jpeg', 'gif'])
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER # UPLOAD FOLDER 英文上传文件夹# 判断上传的文件是否是允许的后缀
def allowed_file(filename):return "." in filename and filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS#把.@app.route('/')
def index():return render_template('upload.html')@app.route('/center/upload',methods=['POST'])#对应表单里面的action="/upload"
def center():if "file" not in request.files: #"file" 对应前端upload.html 中的namereturn redirect(request.url)#redirect 重定向新的地址file = request.files.get('file') # 获取文件if not allowed_file(file.filename): #如果不是传入的指定格式文件。重定向到当前urlreturn redirect(request.url)if file and allowed_file(file.filename):filename = secure_filename(file.filename) # 用这个函数确定文件名称是否是安全file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename)) # 保存文件 os.path.join把目录和文件名合成一个路径return redirect(url_for('show', filename=filename))# 展示
@app.route('/show/<filename>')
def show(filename):# send_from_directory可以从目录加载文件return send_from_directory(app.config['UPLOAD_FOLDER'], filename)if __name__ == '__main__':app.run(host='0.0.0.0', port=5000, debug=True)upload.html
<!DOCTYPE html>
<html lang="en">
<head><meta charset="UTF-8"><meta name="viewport" content="width=device-width, initial-scale=1.0"><title>文件上传</title>
</head>
<body><form action="/center/upload" method="post" enctype="multipart/form-data"><input type="file" name="file"><input type="submit" value="提交"></form></body>
</html>
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)
