当我试图从数据库中获取数据时,我得到的是对象的外键值,而不是值。
在Models.py中
class UP_ROLE_MENU_MAP(models.Model):
iRMID = models.AutoField(primary_key=True)
iRoleID = models.ForeignKey(UP_ROLE_MASTER,on_delete=models.CASCADE,default='',db_column='iRoleID')
iMenuID = models.ForeignKey(up_menus,on_delete=models.CASCADE,default='',db_column='iMenuID')
iStatus = models.PositiveSmallIntegerField(default=1, blank=True,null=True)
在DataTables的视图中
class OrderListJson(BaseDatatableView):
model = UP_ROLE_MENU_MAP
columns = ['iMenuID', 'iDeleteAccess','iAddAccess','iViewAccess']
order_columns = ['iMenuID', 'iDeleteAccess','iAddAccess','iViewAccess']
max_display_length = 10
def render_column(self, row, column):
if column == 'iViewAccess':
return ...
else:
return super(OrderListJson, self).render_column(row, column)
输出:"up_menus object (3)","0","1","link“如何获得值而不是object。
发布于 2018-09-27 14:38:48
要获取id而不是对象,只需将列从
columns = ['iMenuID', 'iDeleteAccess','iAddAccess','iViewAccess']
至
columns = ['iMenuID_id', 'iDeleteAccess','iAddAccess','iViewAccess']
https://stackoverflow.com/questions/52530816
复制相似问题