19  

Your regular expresion has no place holders (that's why you are getting NoReverseMatch):

你的正则表达式的值没有地方持有人(这就是为什么你越来越NoReverseMatch):

url(r'^depict$', cyclops.django.depict, name="cyclops-depict"),

You could do it like this:

你可以这样做:

{% url cyclops-depict %}?smiles=CO&width=200&height=200

URLconf search does not include GET or POST parameters

URLconf搜索不包括GET或POST参数

Or if you wish to use {% url %} tag you should restructure your url pattern to something like

或者，如果您希望使用{% url %}标签，您应该将您的url模式重构为类似的东西。

r'^depict/(?P<width>\d+)/(?P<height>\d+)/(?P<smiles>\w+)$' 

then you could do something like

然后你可以做一些类似的事情

{% url cyclops-depict 200 200 "CO" %}

Follow-up:

追问:

Simple example for custom tag:

自定义标签的简单示例:

from django.core.urlresolvers import reverse
from django import template
register = template.Library()

@register.tag(name="myurl")
def myurl(parser, token):
    tokens = token.split_contents()
    return MyUrlNode(tokens[1:])

class MyUrlNode(template.Node):
    def __init__(self, tokens):
        self.tokens = tokens
    def render(self, context):
        url = reverse('cyclops-depict')
        qs = '&'.join([t for t in self.tokens])
        return '?'.join((url,qs))

You could use this tag in your templates like so:

你可以在你的模板中使用这个标签:

{% myurl width=200 height=200 name=SomeName %}

and hopefully it should output something like

希望它能输出

/depict?width=200&height=200&name=SomeName

43  

Building an url with query string by string concatenation as suggested by some answers is as bad idea as building SQL queries by string concatenation. It is complicated, unelegant and especially dangerous with a user provided (untrusted) input. Unfortunately Django does not offer an easy possibility to pass query parameters to the reverse function.

按照某些答案的建议，通过字符串连接使用查询字符串构建url与通过字符串连接构建SQL查询一样不是个好主意。对于提供(不可信的)输入的用户来说，这是复杂的、不优雅的，尤其危险的。不幸的是，Django并没有提供将查询参数传递给反向函数的简便方法。

Python standard urllib however provides the desired query string encoding functionality.

然而，Python标准urllib提供了所需的查询字符串编码功能。

In my application I've created a helper function:

在我的应用程序中，我创建了一个助手函数:

def url_with_querystring(path, **kwargs):
    return path + '?' + urllib.urlencode(kwargs)

Then I call it in the view as follows:

然后我在视图中这样称呼它:

quick_add_order_url = url_with_querystring(reverse(order_add),
    responsible=employee.id, scheduled_for=datetime.date.today(),
    subject='hello world!')
# http://localhost/myapp/order/add/?responsible=5&
#     scheduled_for=2011-03-17&subject=hello+world%21

Please note the proper encoding of special characters like space and exclamation mark!

请注意空格、感叹号等特殊字符的正确编码!

11  

Neither of the original answers addresses the related issue resolving URLs in view code. For future searchers, if you are trying to do this, use kwargs, something like:

最初的答案都没有解决在视图代码中解决url的相关问题。对于未来的搜索者，如果你想这么做，可以使用kwargs，比如:

reverse('myviewname', kwargs={'pk': value})

反向(myviewname,kwargs = { }“pk”:价值)

9  

I recommend to use builtin django's QueryDict. It also handles lists properly. End automatically escapes some special characters (like =, ?, /, '#'):

我建议使用builtin django的QueryDict。它还能正确地处理列表。End自动转义一些特殊字符(如=、?、/、'#'):

from django.http import QueryDict
from django.core.urlresolvers import reverse

q = QueryDict('', mutable=True)
q['some_key'] = 'some_value'
q.setlist('some_list', [1,2,3])
'%s?%s' % (reverse('some_view_name'), q.urlencode())
# '/some_url/?some_list=1&some_list=2&some_list=3&some_key=some_value'

q.appendlist('some_list', 4)
q['value_with_special_chars'] = 'hello=w#rld?'
'%s?%s' % (reverse('some_view_name'), q.urlencode())
# '/some_url/?value_with_special_chars=hello%3Dw%23rld%3F&some_list=1&some_list=2&some_list=3&some_list=4&some_key=some_value'

To use this in templates you will need to create custom template tag

要在模板中使用它，您需要创建自定义模板标记

5  

The answer that used urllib is indeed good, however while it was trying to avoid strings concatenation, it used it in path + '?' + urllib.urlencode(kwargs). I believe this may create issues when the path has already some query parmas.

使用urllib的答案确实不错，但是当它试图避免字符串连接时，它使用了path + '?' + urllib.urlencode(kwargs)。我认为这可能会在路径已经有一些查询参数时产生问题。

A modified function would look like:

修改后的函数如下:

def url_with_querystring(url, **kwargs):
    url_parts = list(urlparse.urlparse(url))
    query = dict(urlparse.parse_qsl(url_parts[4]))
    query.update(kwargs)
    url_parts[4] = urllib.urlencode(query)
    return urlparse.urlunparse(url_parts)