python flask 上传多个文件，代码怎么写？

Posted 2023-03-19

我已经调出来了服务端接收一个文件和客户端发送一个文件的代码，但是要想接收多个文件和上传多个文件，代码应该怎么改呢？
请大神指点一下。
server.py 的代码如下：
# -*- coding:utf-8 -*-

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

app = Flask(__name__)

app.config['UPLOAD_FOLDER'] = r'E:\\' # UPLOAD_FOLDER 指定了我们文件上传之后在服务器上的存放位置
ALLOWED_EXTENSIONS = set(['txt','pdf','png','jpg','jpeg','gif']) # ALLOWED_EXTENSIONS 则指定了允许上传的文件类型

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

@app.route('/uploadFile', methods = ['GET','POST'])
def upload_file():
if request.method == 'POST':
file = request.files['file']
print file
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
print url_for('uploaded_file', filename=filename)
return "OK"
app.config['MAX_CONTENT_LENGTH'] = 16 * 1024 * 1024

@app.route('/uploaded_file', methods = ['GET','POST'])
def uploaded_file():
request.args.get('filename')

if __name__ == '__main__':
app.run(host='127.0.0.1', port=8972, debug=True)
client.py
代码如下：
# -*- coding:utf-8 -*-

import poster
import cookielib
import urllib2

opener = poster.streaminghttp.register_openers()
opener.add_handler(urllib2.HTTPCookieProcessor(cookielib.CookieJar()))

params = 'file': open(r"F:\test\sql.txt", "rb"), 'name': 'upload_test'
datagen, headers = poster.encode.multipart_encode(params)
request = urllib2.Request('http://127.0.0.1:8972/uploadFile', datagen, headers)
result = urllib2.urlopen(request)
print result

参考技术A include <iostream>
#include <stdio.h>
int jc(int m)
if(m!=1) return m*jc(m-1);
else return 1;

int c(int m,int n)
if(m>=n) return jc(m)/(jc(n)*jc(m-n));

int main(void)
int m,n;
scanf("%d%d",&m,&n);
printf("%d\n",c(m,n));
return 0;
追问

我自己用已经解决了。你这个我看不懂，不过还是要感谢你哈。

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