擅长:python、mysql、java
<p>您不想使用<strong><code>.read()</code></strong>,只需使用<code>filestream</code>所以,请尝试这样的方法<br/></p>
<pre><code>def report_generate(request):
filename = "mysample.zip"
<b>f = open(filename, "rb")</b>
response = HttpResponse(<b>f,</b> content_type='application/x-zip')
response['Content-Disposition'] = 'attachment; filename="%s"' % filename
return response</code></pre>