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django/tests/modeltests/m2m_and_m2o/models.py
Malcolm Tredinnick 953badbea5 Merged Unicode branch into trunk (r4952:5608). This should be fully
backwards compatible for all practical purposes.

Fixed #2391, #2489, #2996, #3322, #3344, #3370, #3406, #3432, #3454, #3492, #3582, #3690, #3878, #3891, #3937, #4039, #4141, #4227, #4286, #4291, #4300, #4452, #4702


git-svn-id: http://code.djangoproject.com/svn/django/trunk@5609 bcc190cf-cafb-0310-a4f2-bffc1f526a37
2007-07-04 12:11:04 +00:00

66 lines
1.6 KiB
Python

"""
29. Many-to-many and many-to-one relationships to the same table
Make sure to set ``related_name`` if you use relationships to the same table.
"""
from django.db import models
class User(models.Model):
username = models.CharField(maxlength=20)
class Issue(models.Model):
num = models.IntegerField()
cc = models.ManyToManyField(User, blank=True, related_name='test_issue_cc')
client = models.ForeignKey(User, related_name='test_issue_client')
def __unicode__(self):
return unicode(self.num)
class Meta:
ordering = ('num',)
__test__ = {'API_TESTS':"""
>>> Issue.objects.all()
[]
>>> r = User(username='russell')
>>> r.save()
>>> g = User(username='gustav')
>>> g.save()
>>> i = Issue(num=1)
>>> i.client = r
>>> i.save()
>>> i2 = Issue(num=2)
>>> i2.client = r
>>> i2.save()
>>> i2.cc.add(r)
>>> i3 = Issue(num=3)
>>> i3.client = g
>>> i3.save()
>>> i3.cc.add(r)
>>> from django.db.models.query import Q
>>> Issue.objects.filter(client=r.id)
[<Issue: 1>, <Issue: 2>]
>>> Issue.objects.filter(client=g.id)
[<Issue: 3>]
>>> Issue.objects.filter(cc__id__exact=g.id)
[]
>>> Issue.objects.filter(cc__id__exact=r.id)
[<Issue: 2>, <Issue: 3>]
# These queries combine results from the m2m and the m2o relationships.
# They're three ways of saying the same thing.
>>> Issue.objects.filter(Q(cc__id__exact=r.id) | Q(client=r.id))
[<Issue: 1>, <Issue: 2>, <Issue: 3>]
>>> Issue.objects.filter(cc__id__exact=r.id) | Issue.objects.filter(client=r.id)
[<Issue: 1>, <Issue: 2>, <Issue: 3>]
>>> Issue.objects.filter(Q(client=r.id) | Q(cc__id__exact=r.id))
[<Issue: 1>, <Issue: 2>, <Issue: 3>]
"""}