Python 实现http server接收mutipart/form-data文件 方法1

• • 1 Server端代码
  • 2 客户端截图
  • 3 代码说明

1 Server端代码

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

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = 'E://recv//'

@app.route('/seenton/monitor/alarmImage', methods=['POST'])
def upload_file():
    head = request.headers
    print(f' recvvvvvvv head= [{head}]')
    content_type = request.content_type
    print(f' recvvvvvvv content_type= [{content_type}]')
    boundary = request.content_type.split(';')[1].split('=')[1]
    print(f' recvvvvvvv boundary= [{boundary}]')
    content_len = request.content_length

    if 'file' not in request.files:
        return 'No file part'
    file = request.files['file']
    if file.filename == '':
        return 'No selected file'

    print(f' file.filename= [{file.filename}]')
    
    if file:
        filename = secure_filename(file.filename)
        file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
        return 'File saved successfully'

if __name__ == '__main__':
    app.run(host='192.168.1.173', port=18098)

2 客户端截图

PostMan生成Python 代码：

import requests

url = "http://192.168.1.16:18098/seenton/monitor/alarmImage"

payload = {}
files=[
  ('file',('测温点位1-58@stator#20220226-221220#stat.jpg',open('/D:/Desktop/20220226-221220-带双光融合效果/测温点位1-58@stator#20220226-221220#stat.jpg','rb'),'image/jpeg'))
]
headers = {}

response = requests.request("POST", url, headers=headers, data=payload, files=files)

print(response.text)

3 代码说明

1. 代码基本功能：实现将客户端发送的文件转发到本地。
2. 代码中的打印：为了调试方便代码中增加了一些关于boudary的打印。
3. 代码中的自定义字段：代码中的 ‘file’ 就是postman客户端请求中的文件名称。