Question

我想通过werkzeug流式传输一个大文件目前我的wsgi应用程序如下所示：

from werkzeug.wrappers import Request, Response
from werkzeug.wsgi import ClosingIterator, wrap_file
import os

class Streamer(object):

    def __init__(self):
        pass

    def __call__(self, environ, start_response):
        request = Request(environ)
        filename = os.getcwd() + "/bigfile.xml"
        try:
            response = wrap_file(environ, open(filename) )
            return response
        except HTTPException, e:
            response = e
            return ClosingIterator(response(environ, start_response))

我不确定我应该对wrap_file函数返回的对象做什么。

Answer 1

没有尝试过，但我认为以下方法可行。

g = file(path_to_bigfile) # or any generator
return Response(g, direct_passthrough=True)

Answer 2

以防万一，另外： 1.保留文件名 2.发出没有页面重定向的下载

# file_name assumed to be known
# file_path assumed to be known
file_size = os.path.getsize(file_path)
fh = file(file_path, 'rb')
return Response(fh,
                mimetype='application/octet-stream',
                headers=[
                    ('Content-Length', str(file_size)),
                    ('Content-Disposition', "attachment; filename=\"%s\"" % file_name),
                ],
                direct_passthrough=True)

如何使用werkzeug流式传输文件？

2 个答案: