Question

我有2张桌子

的OrderItems

+--------+------------+------------------+---------+ | item_id| item_name | Item_type | trans | +--------+------------+------------------+---------+ | 1 | Devon | line_item | 0 | | 2 | 100 | coupon | 0 | | 3 | Contin | line_item | 1 | +--------+------------+------------------+---------+

orderitems_data

1+--------+------------+------------------+-----------+ | id | item_id | Item_key | key_val | +--------+------------+------------------+------------+ | 1 | 1 | Delivery | 2015/04/03 | | 2 | 1 | attrib | pick_up | | 3 | 1 | qty | 1 | | 4 | 2 | discount | 0 | | 5 | 3 | Delivery | 2015/04/15 | | 5 | 3 | attrib | doorstep | +--------+------------+------------------+------------+

我需要从'orderitems'中选择'item_name'，其中'Item_type = line_item'和'trans = 0'，'orderitems_data'中的'key_val'，'item_key = Delivery'，以及'key_val'，其中'item_key = attribute'用于相同的'item_name'

然后显示它们。我正在努力查询，我想写

 SELECT orderitems.item_name AS item_name WHERE orderitems.Item_type='line_item' AND trans=0, orderitems_data.key_val AS deliverydate WHERE Item_key='Delivery', orderitems_data.key_val AS location WHERE Item_key='attrib' FROM orderitems INNER JOIN order_items_data ON orderitems.item_id = orderitems_data.item_id

但显然它不对。

有人能指出我正确的方向吗？

我希望获得的结果是

Devon pick_up 2015/04/03

Answer 1

这会对你有帮助 -

SELECT orderitems.item_name AS item_name, orderitems_data.key_val AS deliverydate, orderitems_data.key_val AS location from orderitems left join orderitems_data
ON orderitems.item_id = orderitems_data.item_id
WHERE orderitems.Item_type='line_item' AND 
    orderitems.trans=0  and 
    ( orderitems_data.Item_key='Delivery' or orderitems_data.Item_key='attrib')

Answer 2

SELECT oi.item_name, oid.key_val
FROM OrderItem oi, Orderitem_Data oid
WHERE oi.item_id = od.item_id
AND oi.item_type = 'line_item'
AND oid.item_key = 'Delivery'

最终没有得到你的要求

Answer 3

我想这会起作用

SELECT i.item_name, d.Item_key FROM orderitems i LEFT JOIN orderitems_data d on i.item_id =d.item_id WHERE i.Item_type='line_item' AND i.trans=0 AND d.Item_key in('attrib', 'Delivery');

Answer 4

这是从EAV数据集构建伪规范化枢轴的股票答案：

SELECT i.*
     , MAX(CASE WHEN d.item_key = 'attrib' THEN key_val END) attrib
     , MAX(CASE WHEN d.item_key = 'Delivery' THEN key_val END) Delivery
     , MAX(CASE WHEN d.item_key = 'qty' THEN key_val END) qty
     , MAX(CASE WHEN d.item_key = 'discount' THEN key_val END) discount
  FROM orderitems i 
  LEFT 
  JOIN orderitems_data d 
    ON d.item_id = i.item_id 
 GROUP 
    BY i.item_id;
+---------+-----------+-----------+-------+----------+------------+------+----------+
| item_id | item_name | Item_type | trans | attrib   | Delivery   | qty  | discount |
+---------+-----------+-----------+-------+----------+------------+------+----------+
|       1 | Devon     | line_item |     0 | pick_up  | 2015/04/03 | 1    | NULL     |
|       2 | 100       | coupon    |     0 | NULL     | NULL       | NULL | 0        |
|       3 | Contin    | line_item |     1 | doorstep | 2015/04/15 | NULL | NULL     |
+---------+-----------+-----------+-------+----------+------------+------+----------+

请注意我之前关于orderitems_data表中id列冗余的注释以及尽可能使用正确数据类型的重要性。

Answer 5

如果，我没有错，您希望获得依赖Item_type = 'line_item'和trans = 0以及od.Item_key = 'Delivery' OR od.Item_key = 'attrib'，然后使用以下查询

SELECT DISTINCT
  o.*,
  od.key_val,
  od.Item_key
FROM
  orderitems o
  LEFT JOIN orderitems_data od
    ON o.item_id = od.item_id 
WHERE o.Item_type = 'line_item' 
  AND o.trans = 0 
  AND (od.Item_key = 'Delivery' OR  od.Item_key = 'attrib')

具有条件的2个表的mysql查询

5 个答案: