如何在django的URL中传递kwargs [英] How can I pass kwargs in URL in django
本文介绍了如何在django的URL中传递kwargs的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在django文档中,url函数就是这样
url(regex,view,kwargs = None,name = None ,prefix ='')
我有这个
url(r'^ download / template /(?P< object_id> \d +)/ $',views.myview()。myfunction,model = models.userModel,name =sample)
这是我的看法
$ my $($)$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ model model model model model model model model model model model model model model model model model model model model model model model model model model model model ']
我收到这个错误
url()得到一个意想不到的关键字参数'model'
解决方案
您正尝试将模型
关键字参数传递给 url()
功能;你需要传递一个 kwargs
参数(它需要一个字典):
url(r'^ download / template /(?P< object_id> \d +)/ $',views.myview()。myfunction,
kwargs = dict(model = models.userModel),name =sample)
In the django doc the url function is like this
url(regex, view, kwargs=None, name=None, prefix='')
I have this
url(r'^download/template/(?P<object_id>\d+)/$', views.myview().myfunction,model=models.userModel, name="sample")
This is my view
class myview(TemplateView):
def myfunction(self,request, object_id, **kwargs):
model = kwargs['model']
I get this error
url() got an unexpected keyword argument 'model'
解决方案
You are trying to pass in a model
keyword argument to the url()
function; you need to pass in a kwargs
argument instead (it takes a dictionary):
url(r'^download/template/(?P<object_id>\d+)/$', views.myview().myfunction,
kwargs=dict(model=models.userModel), name="sample")
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