将大型视频文件上传到Google App Engine

Question

我正在尝试通过谷歌应用引擎将一个大视频上传到谷歌云存储中。

我遵循了这篇文章中的原则，这篇文章提到了上传一个大的电子表格。在GCP应用引擎中无法将大文件上传到Python+Flask

这是我的后端代码

    from flask import Flask, render_template, request, redirect, url_for
    from google.cloud import storage
    from flask_dropzone import Dropzone
    from werkzeug.utils import secure_filename
    import os
    import base64


    app = Flask(__name__, template_folder='./templates', static_folder="./static")
    dropzone = Dropzone(app)
    app.config['UPLOAD_PATH'] = '/tmp'

    @app.route('/', methods=['GET', 'POST'])

    def index():
        return render_template('index.html')


    @app.route('/upload', methods=['POST', 'GET'])
    def upload():

        if request.method == 'POST':

        upload_file = request.files.get('file')

        tmp_file_path = '/tmp/' + upload_file.filename

        with open(tmp_file_path, 'a') as f:
            f.write(uploaded_file.read())

        chunk_index = int(flask.request.form.get('dzchunkindex')) if 
(flask.request.form.get('dzchunkindex') is not None)  else 0
    
        chunk_count = int(flask.request.form.get('dztotalchunkcount')) if (flask.request.form.get('dztotalchunkcount') is not None)  else 1


        if (chunk_index == (chunk_count - 1)):
            print('Saving file to storage')
            print( chunk_count )

            storage_client = storage.Client()

            storage_bucket = storage_client.get_bucket('percy-277618.appspot.com')

            blob = storage_bucket.blob(upload_file.filename)

            blob.upload_from_filename(tmp_file_path, client=storage_client)
            print('Saved to Storage')

            print('Deleting temp file')
            os.remove(tmp_file_path)

                  

if __name__ == '__main__':

    app.run(host='127.0.0.1', port=8080, debug=True)

这是我的前端代码。

Dropzone.options.uploadwidget = {

       paramName: 'file',
       forceChunking: true,
       timeout: 300000,
       chunking: true,
       url: '/upload',
       chunkSize: 10485760,
       maxFilesize: 1025,

};

Answer 1

使用uploaded_file.read()生成的是字节，而不是字符串。您应该以二进制模式打开文件：

with open(tmp_file_path, 'ab') as f: