使用Flask处理大文件上传

小编典典

使用Flask处理大文件上传

flask

用Flask处理超大文件上传（1 GB以上）的最佳方法是什么？

我的应用程序实际上需要多个文件，为它们分配一个唯一的文件号，然后根据用户选择的位置将其保存在服务器上。

我们如何将文件上传作为后台任务运行，以使用户在1小时内没有浏览器旋转，而是可以立即进入下一页？

Flask开发服务器能够处理大量文件（50gb需要1.5个小时，上传速度很快，但将文件写入空白文件的速度却很慢）
如果我用Twisted包装应用程序，则该应用程序在大文件上崩溃
我曾尝试将Celery与Redis结合使用，但这似乎不是发布上传的选项
我在Windows上，网络服务器的选择较少

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2020-04-07

共1个答案

小编典典

我认为解决该问题的超级简单方法只是将文件分成许多小部分/大块发送。因此，要完成这项工作将需要两个部分，即前端（网站）和后端（服务器）。对于前端部分，你可以使用类似的东西Dropzone.js，它没有附加的依赖关系，并且包含不错的CSS。你所要做的就是将类添加dropzone到表单，它会自动将其变成其特殊的拖放字段之一（你也可以单击并选择）。

但是，默认情况下，dropzone不会对文件进行分块。幸运的是，它确实很容易启用。下面是一个示例文件上传形式DropzoneJS和chunking启用：

<html lang="en">
<head>

    <meta charset="UTF-8">

    <link rel="stylesheet" 
     href="https://cdnjs.cloudflare.com/ajax/libs/dropzone/5.4.0/min/dropzone.min.css"/>

    <link rel="stylesheet" 
     href="https://cdnjs.cloudflare.com/ajax/libs/dropzone/5.4.0/min/basic.min.css"/>

    <script type="application/javascript" 
     src="https://cdnjs.cloudflare.com/ajax/libs/dropzone/5.4.0/min/dropzone.min.js">
    </script>

    <title>File Dropper</title>
</head>
<body>

<form method="POST" action='/upload' class="dropzone dz-clickable" 
      id="dropper" enctype="multipart/form-data">
</form>

<script type="application/javascript">
    Dropzone.options.dropper = {
        paramName: 'file',
        chunking: true,
        forceChunking: true,
        url: '/upload',
        maxFilesize: 1025, // megabytes
        chunkSize: 1000000 // bytes
    }
</script>
</body>
</html>

这是使用flask的后端部分：

import logging
import os

from flask import render_template, Blueprint, request, make_response
from werkzeug.utils import secure_filename

from pydrop.config import config

blueprint = Blueprint('templated', __name__, template_folder='templates')

log = logging.getLogger('pydrop')


@blueprint.route('/')
@blueprint.route('/index')
def index():
    # Route to serve the upload form
    return render_template('index.html',
                           page_name='Main',
                           project_name="pydrop")


@blueprint.route('/upload', methods=['POST'])
def upload():
    file = request.files['file']

    save_path = os.path.join(config.data_dir, secure_filename(file.filename))
    current_chunk = int(request.form['dzchunkindex'])

    # If the file already exists it's ok if we are appending to it,
    # but not if it's new file that would overwrite the existing one
    if os.path.exists(save_path) and current_chunk == 0:
        # 400 and 500s will tell dropzone that an error occurred and show an error
        return make_response(('File already exists', 400))

    try:
        with open(save_path, 'ab') as f:
            f.seek(int(request.form['dzchunkbyteoffset']))
            f.write(file.stream.read())
    except OSError:
        # log.exception will include the traceback so we can see what's wrong 
        log.exception('Could not write to file')
        return make_response(("Not sure why,"
                              " but we couldn't write the file to disk", 500))

    total_chunks = int(request.form['dztotalchunkcount'])

    if current_chunk + 1 == total_chunks:
        # This was the last chunk, the file should be complete and the size we expect
        if os.path.getsize(save_path) != int(request.form['dztotalfilesize']):
            log.error(f"File {file.filename} was completed, "
                      f"but has a size mismatch."
                      f"Was {os.path.getsize(save_path)} but we"
                      f" expected {request.form['dztotalfilesize']} ")
            return make_response(('Size mismatch', 500))
        else:
            log.info(f'File {file.filename} has been uploaded successfully')
    else:
        log.debug(f'Chunk {current_chunk + 1} of {total_chunks} '
                  f'for file {file.filename} complete')

    return make_response(("Chunk upload successful", 200))

2020-04-07