python Django 中的值错误
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/2032360/
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
ValueError in Django
提问by Wally
I'm getting a strange error and can't figure out why. I'd appreciate any input. I've been stuck on this for a few days. Here is my code:
我收到一个奇怪的错误，不知道为什么。我很感激任何输入。我已经被困在这几天了。这是我的代码：
models.py
模型.py
class Employee(models.Model):
lastname = models.CharField(max_length=75)
firstname = models.CharField(max_length=75)
position = models.ForeignKey(Position)
jurisdiction = models.ForeignKey(Jurisdiction)
basepay = models.FloatField()
ot = models.FloatField()
benefits = models.FloatField()
totalpay = models.FloatField()
class Meta:
ordering = ['lastname', 'firstname']
def __unicode__(self):
return "%s %s" % (self.firstname, self.lastname)
def full_name(self):
return "%s, %s" % (self.lastname, self.firstname)
def get_absolute_url(self):
return "/salaries/employee/%s/" % self.id
urls.py
网址.py
from django.conf.urls.defaults import *
from djangodemo.salaries.models import Employee
from django.views.generic import list_detail
employee_info = {
"queryset" : Employee.objects.all(),
"template_name" : "salaries/employee.html",
}
urlpatterns = patterns('',
(r'^salaries/employee/$', list_detail.object_list, 'employee_info'),
)
employee.html
员工.html
{{ object_list }}
When I run python manage.py runserver and look at http://127.0.0.1:8000/salaries/employeein my browser, I get this error:
当我运行 python manage.py runserver 并http://127.0.0.1:8000/salaries/employee在浏览器中查看时，出现此错误：
Traceback (most recent call last):
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 279, in run
self.result = application(self.environ, self.start_response)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\servers\basehttp.py", line 651, in __call__
return self.application(environ, start_response)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\handlers\wsgi.py", line 241, in __call__
response = self.get_response(request)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\handlers\base.py", line 73, in get_response
response = middleware_method(request)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\middleware\common.py", line 57, in process_request
_is_valid_path("%s/" % request.path_info)):
File "F:\django\instantdjango\Python26\Lib\site-packages\django\middleware\common.py", line 142, in _is_valid_path
urlresolvers.resolve(path)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\urlresolvers.py", line 294, in resolve
return get_resolver(urlconf).resolve(path)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\urlresolvers.py", line 218, in resolve
sub_match = pattern.resolve(new_path)
File "F:\django\instantdjango\Python26\Lib\site-packages\django\core\urlresolvers.py", line 123, in resolve
kwargs.update(self.default_args)
ValueError: dictionary update sequence element #0 has length 1; 2 is required
回答by Roy Tang
urlpatterns = patterns('',
(r'^salaries/employee/$', list_detail.object_list, 'employee_info'),
)
The third item in the tuple needs to be a dictionary, not a string. Try removing the single quotes around employee_info:
元组中的第三项需要是字典，而不是字符串。尝试删除employee_info 周围的单引号：
urlpatterns = patterns('',
(r'^salaries/employee/$', list_detail.object_list, employee_info),
)
回答by panchicore
may be you mean the URL name:
可能是您的意思是网址name：
urlpatterns = patterns('',
(r'^salaries/employee/$', list_detail.object_list, name='employee_info'),
)
