如何将绝对 URL 传递给 django 函数并重定向到它?
def function(request):
back_url = request.META['HTTP_REFERER'] # example.com/home/?status=80&page=1
return redirect(back_url)
最佳答案
你可以这样做:
def func(request):
url = request.META['HTTP_REFERER']
if request.META['QUERY_STRING']:
url += '?%s' % request.META['QUERY_STRING']
return redirect_to(request, url, **kwargs)
关于python - 如何在django中传递绝对url,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16598292/