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我有以下带有一些示例数据的 mysql 表。

1) 表：ai_shipment

+----+---------------------+---------------------+------------------+
| id | booking_date        | loading_date        | container_number |
+----+---------------------+---------------------+------------------+
|  1 | 2012-08-03 00:00:00 | 2012-08-04 00:00:00 | ABB987987BBC6    |
|  2 | 2012-08-05 00:00:00 | 2012-08-07 00:00:00 | BHJKKU78786GH    |
+----+---------------------+---------------------+------------------+

2）表：ai_purchase_item 

+----+---------+----------+-----------+-----------+
| id | item_id | quantity | cost      | rate      |
+----+---------+----------+-----------+-----------+
|  1 |       1 |       50 | 1200.0000 | 1355.0000 |
|  2 |       2 |       20 |  550.0000 |  675.0000 |
|  3 |       4 |       70 |   70.0000 |   70.0000 |
|  4 |       6 |       90 |   90.0000 |   90.0000 |
|  5 |       7 |       80 |   80.0000 |   80.0000 |
+----+---------+----------+-----------+-----------+

3) 表：shipping_purchase_item

+----+-------------+------------------+
| id | shipment_id | purchase_item_id |
+----+-------------+------------------+
|  1 |           1 |                2 |
|  2 |           2 |                3 |
+----+-------------+------------------+

基本上，我将所有购买的物品详细信息存储在 中ai_purchase_item，ai_shipment存储运输详细信息，并ai_shipment_purchase_item存储已运输物品的记录。

我想做的是，我想从中选择所有ai_purchase_item未发货的记录。这意味着外键不应该存在于ai_shipment_purchase_item.

参考上面的记录，我期望得到的结果是。

+----+---------+----------+-----------+-----------+
| id | item_id | quantity | cost      | rate      |
+----+---------+----------+-----------+-----------+
|  1 |       1 |       50 | 1200.0000 | 1355.0000 |
|  4 |       6 |       90 |   90.0000 |   90.0000 |
|  5 |       7 |       80 |   80.0000 |   80.0000 |
+----+---------+----------+-----------+-----------+

我尝试过这样的事情（我知道 sql 查询不正确）

 SELECT 
    pi.id, 
    pi.item_id, 
    pi.quantity, 
    pi.cost,
    pi.rate 
 FROM 
    ai_purchase_item pi 
 JOIN 
    ai_shipment_purchase_item spi ON spi.purchase_item_id = pi.id
 WHERE NOT EXISTS (SELECT spi.purchase_item_id WHERE spi.purchase_item_id = pi.id)

谢谢你。

4

2 回答 2

3

在 MySQL 中，有三种常见的方法来查找一个表中存在但在另一个表中缺失的值：

  • LEFT JOIN
  • NOT EXISTS
  • NOT IN

这是使用的方法NOT EXISTS：

SELECT 
    pi.id, 
    pi.item_id, 
    pi.quantity, 
    pi.cost,
    pi.rate 
FROM 
    ai_purchase_item AS pi 
WHERE NOT EXISTS
(
    SELECT *
    FROM ai_shipment_purchase_item AS spi
    WHERE spi.purchase_item_id = pi.id
)

这是一个LEFT JOIN：

 SELECT 
    pi.id, 
    pi.item_id, 
    pi.quantity, 
    pi.cost,
    pi.rate 
 FROM 
    ai_purchase_item AS pi 
 LEFT JOIN 
    ai_shipment_purchase_item AS spi ON spi.purchase_item_id = pi.id
 WHERE spi.purchase_item_id IS NULL

这是NOT IN：

SELECT 
    pi.id, 
    pi.item_id, 
    pi.quantity, 
    pi.cost,
    pi.rate 
FROM 
    ai_purchase_item AS pi 
WHERE pi.id NOT IN
(
     SELECT purchase_item_id
     FROM ai_shipment_purchase_item
)

Quassnoi 的以下文章比较了每种方法的性能：

结论是：

在 MySQL 中搜索缺失值的最佳方法是使用 LEFT JOIN / IS NULL 或 NOT IN 而不是 NOT EXISTS。

于 2012-07-30T10:40:05.107 回答
1 
select * from ai_purchase_item where id not in (select distinct(purchase_item_id) from shipment_purchase_item )
于 2012-07-30T10:41:53.440 回答