Question

我想在contains查询中使用join，但它会出错。

以下是我的 models.py ：

 class Profile(models.Model):
     name = models.CharField(max_length=50, primary_key=True)
     assign = models.CharField(max_length=50)
     doj = models.DateField()
     dob = models.DateField()

     class Meta:
        db_table = u'profile'

     def __str__(self):
         return  '%s %s %s %s' % ( self.name,self.assign,self.doj,self.dob)

     def __unicode__(self):
         return  u'%s %s %s %s' % ( self.name,self.assign,self.doj,self.dob)


    enter code here

class working(models.Model):
   w_name =models.ForeignKey(Profile, db_column='w_name')
   monday =  models.IntegerField(null=True, db_column='monday', blank=True)
   tuesday =  models.IntegerField(null=True, db_column='tuesday', blank=True)
   wednesday =  models.IntegerField(null=True, db_column='wednesday', blank=True)

   class Meta:
        db_table = u'working'

   def __str__(self):
         return  '%s %s %s %s' % ( self.w_name,self.monday,self.tuesday,self.wednesday)

   def __unicode__(self):
         return  u'%s %s %s %s' % ( self.w_name,self.monday,self.tuesday,self.wednesday)

我想使用这样的东西：

m = working.objects.filter(w_name__name__contains='sushanth')

or 

m = working.objects.filter(w_name__name__startswith='sushanth')

我一直得到的错误：

     File "<console>", line 1, in <module>
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/manager.py", line 141, in filter
    return self.get_query_set().filter(*args, **kwargs)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/query.py", line 556, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/query.py", line 574, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/sql/query.py", line 1152, in add_q
    can_reuse=used_aliases)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/sql/query.py", line 1092, in add_filter
    connector)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/sql/where.py", line 67, in add
    value = obj.prepare(lookup_type, value)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/sql/where.py", line 316, in prepare
    return self.field.get_prep_lookup(lookup_type, value)
  File "/usr/local/lib/python2.6/dist-packages/Django-1.2.4-py2.6.egg/django/db/models/fields/related.py", line 139, in get_prep_lookup
    raise TypeError("Related Field has invalid lookup: %s" % lookup_type)
TypeError: Related Field has invalid lookup: startswith

有什么建议吗？

Answer 1

名称字段是主键有什么原因吗？

如果没有，那么我建议使用名称字段NOT作为主键重新创建表。这可能会抛弃过滤器逻辑。

如何在连接查询中使用contains或startswith

1 个答案: