我想将项目目录中的PDF文件设为可下载,而不是在用户点击链接时在浏览器中打开。
我关注了这个问题Generating file to download with Django
但我收到错误:
Exception Type: SyntaxError
Exception Value: can't assign to literal (views.py, line 119)
Exception Location: /usr/local/lib/python2.7/dist-packages/django/utils/importlib.py in import_module, line 35
我创建了一个下载链接:
<a href="/files/pdf/resume.pdf" target="_blank" class="btn btn-success btn-download" id="download" >Download PDF</a>
urls.py:
url(r'^files/pdf/(?P<filename>\{w{40})/$', 'github.views.pdf_download'),
views.py:
def pdf_download(request, filename):
path = os.expanduser('~/files/pdf/')
f = open(path+filename, "r")
response = HttpResponse(FileWrapper(f), content_type='application/pdf')
response = ['Content-Disposition'] = 'attachment; filename=resume.pdf'
f.close()
return response
错误行是:
response = ['Content-Disposition'] = 'attachment; filename=resume.pdf'
如何让它可以下载?
谢谢!
更新
它适用于Firefox,但不适用于Chrome v21.0。
答案 0 :(得分:5)
你在该行中有一个额外的=
,这会使语法无效。它应该是
response['Content-Disposition'] = 'attachment; filename=resume.pdf'
(请注意,有两个=
并不一定会使其无效:foo = bar = 'hello'
完全有效,但在这种情况下,左侧和中间两个名称都是名称。在您的版本中,中间term是一个文字,不能分配给。)
答案 1 :(得分:4)
使用以下代码,它应该下载文件而不是在新页面中打开它
def pdf_download(request, filename):
path = os.expanduser('~/files/pdf/')
wrapper = FileWrapper(file(filename,'rb'))
response = HttpResponse(wrapper, content_type=mimetypes.guess_type(filename)[0])
response['Content-Length'] = os.path.getsize(filename)
response['Content-Disposition'] = "attachment; filename=" + filename
return response