如何使用pyramid.response.FileIter

Question

我有以下视图代码,试图将zip文件"流"到客户端进行下载:

import os
import zipfile
import tempfile
from pyramid.response import FileIter

def zipper(request):
    _temp_path = request.registry.settings['_temp']
    tmpfile = tempfile.NamedTemporaryFile('w', dir=_temp_path, delete=True)

    tmpfile_path = tmpfile.name

    ## creating zipfile and adding files
    z = zipfile.ZipFile(tmpfile_path, "w")
    z.write('somefile1.txt')
    z.write('somefile2.txt')
    z.close()

    ## renaming the zipfile
    new_zip_path = _temp_path + '/somefilegroup.zip'
    os.rename(tmpfile_path, new_zip_path)

    ## re-opening the zipfile with new name
    z = zipfile.ZipFile(new_zip_path, 'r')
    response = FileIter(z.fp)

    return response

但是,这是我在浏览器中获得的响应:

Could not convert return value of the view callable function newsite.static.zipper into a response object. The value returned was .

我想我没有正确使用FileIter.

---

更新:

自从使用Michael Merickel的建议更新后,FileIter函数正常工作.但是,仍然挥之不去的是客户端(浏览器)上出现的MIME类型错误: Resource interpreted as Document but transferred with MIME type application/zip: "http://newsite.local:6543/zipper?data=%7B%22ids%22%3A%5B6%2C7%5D%7D"

为了更好地说明这个问题,我已经包括一个微小.py和.pt文件Github上:https://github.com/thapar/zipper-fix

Answer 1

FileIter不是响应对象,就像您的错误消息所说的那样.它是一个可以用于响应体的迭代,就是这样.此外,它ZipFile可以接受文件对象,这比文件路径更有用.让我们尝试写入tmpfile,然后将该文件指针倒回到开头,并使用它写出而不做任何花哨的重命名.

import os
import zipfile
import tempfile
from pyramid.response import FileIter

def zipper(request):
    _temp_path = request.registry.settings['_temp']
    fp = tempfile.NamedTemporaryFile('w+b', dir=_temp_path, delete=True)

    ## creating zipfile and adding files
    z = zipfile.ZipFile(fp, "w")
    z.write('somefile1.txt')
    z.write('somefile2.txt')
    z.close()

    # rewind fp back to start of the file
    fp.seek(0)

    response = request.response
    response.content_type = 'application/zip'
    response.app_iter = FileIter(fp)
    return response

我根据文档更改了模式,NamedTemporaryFile以'w+b'允许写入和读取文件.