使用 Flask 处理大文件上传

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使用 Flask 处理超大文件上传(1 GB 以上)的最佳方式是什么?

我的应用程序本质上是获取多个文件,为它们分配一个唯一的文件编号,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,以便用户不会让浏览器旋转 1 小时,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 小时,上传很快但将文件写入空白文件非常慢)
  • 如果我用 Twisted 包装应用程序,应用程序会在处理大文件时崩溃
  • 我试过将 Celery 与 Redis 一起使用,但这似乎不是发布上传的选项
  • 我在 Windows 上,网络服务器的选项较少

原文由 Infinity8 发布,翻译遵循 CC BY-SA 4.0 许可协议

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1 个回答

我认为解决问题的超级简单方法就是将文件分成许多小部分/块。因此,将有两个部分来完成这项工作,即前端(网站)和后端(服务器)。对于前端部分,您可以使用类似 Dropzone.js 类的东西,它没有额外的依赖项并且包含不错的 CSS。您所要做的就是将类 dropzone 添加到一个表单中,它会自动将其变成它们的特殊拖放字段之一(您也可以单击并选择)。

但是,默认情况下,dropzone 不会分块文件。幸运的是,启用它真的很容易。这是启用了 DropzoneJSchunking 的示例文件上传表单:

 <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>

这是使用烧瓶的后端部分:

 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))

原文由 Abdul Rehman 发布,翻译遵循 CC BY-SA 4.0 许可协议

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