如何将字典列表转换为字典.下面是dict列表
data = [{'name': 'John Doe', 'age': 37, 'sex': 'M'},
{'name': 'Lisa Simpson', 'age': 17, 'sex': 'F'},
{'name': 'Bill Clinton', 'age': 57, 'sex': 'M'}]
至
data = {{'name': 'John Doe', 'age': 37, 'sex': 'M'},
{'name': 'Lisa Simpson', 'age': 17, 'sex': 'F'},
{'name': 'Bill Clinton', 'age': 57, 'sex': 'M'}}
我正在尝试创建一个表单,在ModelChoiceField从queryset加载,我想向ModelChoiceField添加一些自定义值,我已经使用了选择字段,如下所示,但在更新表单时,得到以下错误
表单错误:选择有效选项.这种选择不是可用的选择之一.
代码:
self.fields['lead'] = forms.ModelChoiceField(queryset = Pepole.objects.filter(poc__in = ('lead','sr.lead')))
self.fields['lead2'] = forms.ModelChoiceField(queryset = Pepole.objects.filter(role__in = ('lead','sr.lead')))
choice_field = self.fields['lead']
choice_field.choices = list(choice_field.choices) + [('None', 'None')]
choice_field = self.fields['lead2']
choice_field.choices = list(choice_field.choices) + [('None', 'None')]
我在这做错了吗？
我试图从DJango中两个相关表的连接中进行选择,如下所示.但我无法得到name另一张桌子的领域.
在SQL中我们可以写:
select person.phonenumber,membership.* from membership where person=name ;
这会显示所有membership字段phonenumber,但在Django中会抛出错误:
无法将关键字"phonenumber"解析为字段
我甚至试过select_related但没用过.我做错了吗？
model.py:
from django.db import models
# Create your models here.
class Person(models.Model):
name = models.CharField(primary_key=True,max_length=128)
phonenumber = models.CharField(max_length=128)
city = models.CharField(max_length=128)
def __unicode__(self):
return self.name
class Membership(models.Model):
person = models.ForeignKey(Person)
date_joined = models.DateField()
invite_reason = models.CharField(max_length=64)
>>> print m.query
SELECT `myapp_membership`.`id`, `myapp_membership`.`person_id`, `myapp_membership`.`date_joined`, `myapp_membership`.`invite_reason` FROM `myapp_membership` WHERE `myapp_membership`.`person_id` = x
>>> m=Membership.objects.filter(person__name='x').select_related()
>>> print m.query
SELECT `myapp_membership`.`id`, `myapp_membership`.`person_id`, `myapp_membership`.`date_joined`, `myapp_membership`.`invite_reason`, …如何在日期范围上获取模型过滤器.在我使用员工doj的项目中,我需要设计一个表.喜欢加入不到三个月,3-6个月,6-12个月,12-24个月的员工.
Depart < 3month 3-6months 6-12months 12-24months
----------------------------------------- ---- -----
A dep 4 6 6 8
------------------------------------------------------
如何在django中执行此过滤器.
我已经通过这个链接,但它令人困惑.
http://www.nerdydork.com/django-filter-model-on-date-range.html
提前致谢
如何在django查询集中为列名写别名.
例如在mysql中:
select m as n, b as a from xyz
我怎么能在django查询集中这样做？
models.Table.objects.all().values('m', 'b')
任何帮助真的很感激.
从1.7升级到1.8后,我面临着一个Django ValidationError.我搜索了这个错误,但没有运气.
在https://github.com/django/django/blob/master/django%2Fdb%2Fmodels%2Ffields%2F__init__.py#L958引发错误
任何帮助真的很感激.
Django Version: 1.8.3
Exception Type: ValidationError
Exception Value:
[u"'xyz' value must be an integer."]
追溯
Traceback:
File "django/v1_8/core/handlers/base.py" in get_response
116. response = middleware_method(request)
File "/local_middleware.py" in process_request
49. admin = request.user.groups.filter(name='admin')
File "django/v1_8/utils/functional.py" in inner
226. self._setup()
File "django/v1_8/utils/functional.py" in _setup
359. self._wrapped = self._setupfunc()
File "django/v1_8/contrib/auth/middleware.py" in <lambda>
22. request.user = SimpleLazyObject(lambda: get_user(request))
File "django/v1_8/contrib/auth/middleware.py" in get_user
10. request._cached_user = auth.get_user(request)
File "django/v1_8/contrib/auth/__init__.py" in get_user
167. user_id = _get_user_session_key(request)
File "django/v1_8/contrib/auth/__init__.py" in …我正在尝试创建一个动态过滤的下拉选项字段,我在博客下面经历过但令人困惑,任何人都可以建议在django中执行此操作.
我正在尝试在Django中创建一个动态过滤的下拉选择字段.我已经尝试了这里概述的步骤,但我不明白.
如何使用ajax在Django中创建过滤的Dropdown Choice字段？
我想在HTML中求和,但模板标签返回0,
View.py
def gen_Report(request):
### query returns below output
list=[{'total': 1744, 'user': u'x'}, {'total': 13, 'user': u'y'}, {'total': 126, 'user': u'z'}, {'total': 46, 'user': u'm'}, {'total': 4, 'user': u'n'}, {'total': 8, 'user': u'o'}, {'total': 3, 'user': u'p'}]
return render_to_response('user.html', locals(),
context_instance = RequestContext(request))
模板:
user.html
{% load temptags %}
<table id="myTable" class="tablesorter">
<thead>
<tr>
<th>S.No</th>
<th>role</th>
<th>Count</th>
</tr>
</thead>
{% for fetch in list %}
<tr>
<td>{{forloop.counter}}</td>
<td>{{fetch.user}}</td>
<td>{{fetch.total}}</td>
{% endfor %}
<td>{{ list.total|running_total}}</td>
<tr>
</table>
模板标签:
from django.template import …如何在下表输出格式中写下面的数据？
字典列表
data=[{'date': datetime.date(2011, 2, 8), 'user': u'xxx', 'status': u'P'}, {'date': datetime.date(2011, 2, 8), 'user': u'yyy', 'status': u'P'}, {'date': datetime.date(2011, 2, 8), 'user': u'zzz', 'status': u'P'}, {'date': datetime.date(2011, 2, 9), 'user': u'xxx, 'status': u'P'}, {'date': datetime.date(2011, 2, 9), 'user': u'yyy', 'status': u'E'}, {'date': datetime.date(2011, 2, 9), 'user': u'zzz', 'status': u'E'}, {'date': datetime.date(2011, 2, 10), 'user': u'xxx', 'status': u'P'}, {'date': datetime.date(2011, 2, 10), 'user': u'yyy', 'status': u'P'}, {'date': datetime.date(2011, 2, 10), 'user': u'zzz', 'status': u'P'}]
输出格式应为:
S.no user 2011-02-08 2011-02-09 2011-02-10 …下面是使用尝试生成dict的dict列表
data = [
{'date': datetime.date(2011, 8, 14), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 15), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 16), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 17), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 18), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 19), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 20), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 21), 'name': u'ab', 'total': 8},
{'date': datetime.date(2011, 8, 18), 'name': u'aj', 'total': 0},
{'date': datetime.date(2011, 8, 14), …