【发布时间】:2017-11-16 20:41:45
【问题描述】:
我有两张桌子
front_employee(Django 中的员工模型)
+-----------------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| order | int(11) | NO | | NULL | |
| name | varchar(100) | YES | | NULL | |
| position | varchar(100) | YES | | NULL | |
| description | longtext | YES | | NULL | |
| employee_img_id | int(11) | NO | MUL | NULL | |
| language_id | int(11) | NO | MUL | NULL | |
+-----------------+--------------+------+-----+---------+----------------+
和front_employeepicture(Django中的EmployeePicture)
+-------+--------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------+--------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| order | int(11) | NO | | NULL | |
| img | varchar(100) | YES | | NULL | |
+-------+--------------+------+-----+---------+----------------+
我想执行这个查询:
SELECT a.id, a.name, b.img
FROM front_employee a
INNER JOIN front_employeepicture b
ON a.employee_img_id = b.id
现在我有
context['employee'] = Employee.objects.all().order_by('order')
我尝试了类似的东西
context['employee'] = Employee.objects.select_related('EmployeePicture')
没有结果。有什么想法吗?
【问题讨论】:
-
您需要展示您的模型。通常,Django 通过
ForeignKey字段为您处理内部连接 - 如果您正确设置模型,这将是微不足道的。