Python 使用django filter进行过滤时,如何将过滤后的值导出到csv文件中
我使用了django过滤模块进行过滤。要将结果导出到CSV文件中Python 使用django filter进行过滤时,如何将过滤后的值导出到csv文件中,python,django,export-to-csv,django-filter,Python,Django,Export To Csv,Django Filter,我使用了django过滤模块进行过滤。要将结果导出到CSV文件中 def bfs_version_filter(request): version_obj = bfs_versions.objects.all() filter_obj = version_filter(request.GET, queryset = version_obj) response = HttpResponse(content_type = 'text/csv') file_name =
def bfs_version_filter(request):
version_obj = bfs_versions.objects.all()
filter_obj = version_filter(request.GET, queryset = version_obj)
response = HttpResponse(content_type = 'text/csv')
file_name = "version_filter"+str(date.today())+".csv"
response['Content-Disposition'] = 'attachment; filename = "'+ file_name +'"' #edited by vennilam
writer = csv.writer(response)
for i in filter_obj:
writer.writerow(i)
return response
获取以下错误:
/bfslite/version\u过滤器处的类型错误/
“version\u filter”对象不可编辑调用
过滤器集
子类的构造函数不会过滤查询集,它将构造该子类的对象
您可以访问以访问筛选的查询集:
def bfs_version_filter(request):
version_obj = bfs_versions.objects.all()
filter_obj = version_filter(request.GET, queryset = version_obj).qs
response = HttpResponse(content_type = 'text/csv')
file_name = "version_filter"+str(date.today())+".csv"
response['Content-Disposition'] = 'attachment; filename = "'+ file_name +'"' #edited by vennilam
writer = csv.writer(response)
for i in filter_obj:
writer.writerow(i)
return response
def bfs\u版本过滤器(请求):
version_obj=bfs_versions.objects.all()
filter\u obj=version\u filter(request.GET,queryset=version\u obj).qs
response=HttpResponse(内容类型='text/csv')
file_name=“version_filter”+str(date.today())+“.csv”
响应['Content-Disposition']='附件;filename=“'+文件名+”#由vennilam编辑
writer=csv.writer(响应)
对于过滤器中的i_obj:
writer.writerow(一)
返回响应
什么是version\u filter()
?您不应该访问.qs
,所以version\u filter(…).qs
?