创建内存中的zip文件并作为http响应返回的功能 [英] Function to create in-memory zip file and return as http response

问题描述

我正在避免在磁盘上创建文件，这是我到目前为止所获得的:

I am avoiding the creation of files on disk, this is what I have got so far:

def get_zip(request):
    import zipfile, StringIO
    i = open('picture.jpg', 'rb').read()
    o = StringIO.StringIO()
    zf = zipfile.ZipFile(o, mode='w')
    zf.writestr('picture.jpg', i)
    zf.close()
    o.seek(0)
    response = HttpResponse(o.read())
    o.close()
    response['Content-Type'] = 'application/octet-stream'
    response['Content-Disposition'] = "attachment; filename=\"picture.zip\""
    return response

您认为正确优雅的Python语言够用吗?还有更好的方法吗?

Do you think is correct-elegant-pythonic enough? Any better way to do it?

谢谢！

推荐答案

对于StringIO，通常应使用o.getvalue()来获取结果.另外，如果要将普通文件添加到zip文件中，则可以使用zf.write('picture.jpg').您不需要手动阅读它.

For StringIO you should generally use o.getvalue() to get the result. Also, if you want to add a normal file to the zip file, you can use zf.write('picture.jpg'). You don't need to manually read it.

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