Python 如何在 django 中通过 url 传递参数？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/32567808/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How do I pass parameters via url in django?
提问by Sa?a Kalaba
I am trying to pass a parameter to my view, but I keep getting this error:
我正在尝试将参数传递给我的视图，但我不断收到此错误：
NoReverseMatch at /pay/how
Reverse for 'pay_summary' with arguments '(False,)' and keyword arguments '{}' not found. 1 pattern(s) tried: ['pay/summary/$']
/pay/howis the current view that I'm at. (that is the current template that that view is returning).
/pay/how是我目前的观点。（即该视图返回的当前模板）。
urls.py
网址.py
url(r'^pay/summary/$', views.pay_summary, name='pay_summary')
views.py
视图.py
def pay_summary(req, option):
if option:
#do something
else:
#do something else
....
template
模板
<a href="{% url 'pay_summary' False %}">my link</a>
EDIT
编辑
I want the view should accept a POST request, not GET.
我希望视图应该接受 POST 请求，而不是 GET。
采纳答案by Dric512
You need to define a variable on the url. For example:
您需要在 url 上定义一个变量。例如：
url(r'^pay/summary/(?P<value>\d+)/$', views.pay_summary, name='pay_summary')),
In this case you would be able to call pay/summary/0
在这种情况下，您可以调用 pay/summary/0
It could be a string true/false by replacing \d+to \s+, but you would need to interpret the string, which is not the best.
通过替换\d+to \s+，它可能是一个字符串真/假，但您需要解释该字符串，这不是最好的。
You can then use:
然后您可以使用：
<a href="{% url 'pay_summary' value=0 %}">my link</a>
回答by Rexcirus
To add to the accepted answer, in Django 2.0 the url syntax has changed:
要添加到已接受的答案中，在 Django 2.0 中，url 语法已更改：
path('<int:key_id>/', views.myview, name='myname')
Or with regular expressions:
或者使用正则表达式：
re_path(r'^(?P<key_id>[0-9])/$', views.myview, name='myname')
