python request向服务端发送文件

本篇文章主要介绍1. 如何使用python request向服务端发送文件 2. 服务端如何接收文件 3. 服务端如何发送文件

1. 如何使用python request向服务端发送文件
   request.post可以发送file类型

def foo():
    local_url = "http://127.0.0.1:5000/test_api"
    file = {'video': open("/path/to/video.mp4", 'rb')}

    res = requests.post(url=local_url,
                        files=file,
                        data={"filename": "video.mp4"},
                        timeout=100000)

2. 服务端如何接收文件
   服务端使用flask搭建，可以接受file类型

from flask import Flask, request
app = Flask(__name__)
@app.route("/test_api", methods=['POST'])
def entrypoint():
    # create cache folder
    save_folder = 'tmp'
    if os.path.exists(save_folder):
        shutil.rmtree(save_folder)
    os.mkdir(save_folder)
    # get video
    video = request.files['video']
    filename = request.form['filename']
    # save video
    video_load_path = os.path.join(save_folder, filename)
    video.save(video_load_path)

if __name__ == '__main__':
    app.run(host="0.0.0.0", port=5000)

3. 服务端发送文件
   基于flask搭建的服务可以使用send_file() 返回文件

from flask import Flask, request
app = Flask(__name__)
@app.route("/test_api", methods=['POST'])
def entrypoint():
    # create cache folder
    save_folder = 'tmp'
    if os.path.exists(save_folder):
        shutil.rmtree(save_folder)
    os.mkdir(save_folder)
    # get video
    video = request.files['video']
    filename = request.form['filename']
    # save video
    video_load_path = os.path.join(save_folder, filename)
    video.save(video_load_path)

    # send file to client
    return send_file(video_load_path, mimetype='video/mp4', as_attachment=True)

if __name__ == '__main__':
    app.run(host="0.0.0.0", port=5000)

客户端接收服务端的发送

def foo():
    local_url = "http://127.0.0.1:5000/test_api"
    file = {'video': open("/path/to/video.mp4", 'rb')}

    res = requests.post(url=local_url,
                        files=file,
                        data={"filename": "video.mp4"},
                        timeout=100000)
    with open("response.mp4", 'wb') as f:
        f.write(res.content)

或者也可以把文件通过cdn发送到用于存储文件的地方，并将存储地址返回给用户，用户通过此地址访问文件

VIDEO_UPLOAD_URL = 'http://xxxxx/upload_video.php'

def upload_video(video_bytes):
    files = {
        'file': video_bytes
    }

    try:
        resp = requests.post(VIDEO_UPLOAD_URL, files=files)
        if resp.status_code == 200:
            url = json.loads(resp.text)['url']
            crc = binascii.crc32(video_bytes)
            url = '{}?crc={}&type=5'.format(url, crc)
            return url
        else:
            return None
    except Exception as err:
        logging.error('upload_video failed, error info {}'.format(err))
        return None

from flask import Flask, request
app = Flask(__name__)
@app.route("/test_api", methods=['POST'])
def entrypoint():
    # create cache folder
    save_folder = 'tmp'
    if os.path.exists(save_folder):
        shutil.rmtree(save_folder)
    os.mkdir(save_folder)
    # get video
    video = request.files['video']
    filename = request.form['filename']
    # save video
    video_load_path = os.path.join(save_folder, filename)
    video.save(video_load_path)
    
    # upload video
    video_url = upload_video(open(video_load_path, 'rb').read()) 
    return jsonify({"status": "success", "url": video_url})

if __name__ == '__main__':
    app.run(host="0.0.0.0", port=5000)