Question

我有两张桌子，其中包含顾客的年龄和身高。

Table: Ages
+-----------+------------+
|customerId |     age    |
+-----------+------------+
|     1     |     15     |
|     2     |     24     |
|     3     |     21     |
|     4     |     62     |
|     6     |     57     |
|     7     |     32     |
+-----------+------------+

Table: Heights
+-----------+------------+
|customerId |   height   |
+-----------+------------+
|     1     |     175    |
|     2     |     182    |
|     4     |     180    |
|     5     |     171    |
|     6     |     165    |
|     7     |     182    |
+-----------+------------+

我需要编写一个SELECT查询来读取所有年龄和高度。所以像这样......

SELECT Ages.age, Heights.height 
FROM Ages INNER JOIN Heights ON Ages.customerId=Heights.customerId;

然而（这里是扭曲）由于记录保持不稳定，两个表都有丢失的记录。（例如，年龄为customerId 5，高度为customerId 3）。

有没有办法编写查询以便它仍然可以工作，但只要数据丢失就返回零？

即

+-----------+------------+------------+
|customerId |     age    |   height   |
+-----------+------------+------------+
|     1     |     15     |     175    |
|     2     |     24     |     182    |
|     3     |     21     |     0      |
|     4     |     62     |     180    |
|     5     |     0      |     171    |
|     6     |     57     |     165    |
|     7     |     32     |     182    |
+-----------+------------+------------+

Answer 1

一种方法（他们是其他人，一如既往）

select customerId, max(age), max(height)
from 
 (
  select customerId, age, 0 as height from Ages
  UNION
  select customerId, 0 as age, height from heights
 ) s
group by customerId;

请参阅SqlFiddle

Answer 2

MySql没有完整的外部联接，但您可以simulate one使用LEFT JOIN，然后使用RIGHT JOIN，并结合UNION，这将结合使用+消除重复：

SELECT Ages.age, COALESCE(Heights.height, 0)
FROM Ages 
LEFT OUTER JOIN Heights ON Ages.customerId=Heights.customerId
UNION
SELECT COALESCE(Ages.age, 0), Heights.height 
FROM Ages 
RIGHT OUTER JOIN Heights ON Ages.customerId=Heights.customerId;

SqlFiddle Here

Answer 3

你真正需要的是full outer join，但MySQL并不支持。相反，让所有客户都在子查询中并使用left outer join：

select c.customerid, coalesce(a.age, 0) as age, coalesce(h.height, 0) as height
from (select customerid from ages union
      select customerid from heights
     ) c left outer join
     ages a
     on a.customerid = c.customerid left outer join
     heights h
     on h.customerid = c.customerid;

Answer 4

~~使用LEFT JOIN，IF条件为NULL~~

~~SELECT Ages.age, IF (Heights.height IS NULL, 0, Heights.height) AS height FROM Ages LEFT JOIN Heights ON Ages.customerId=Heights.customerId;~~

~~好的，快速回答......上面只会给你0作为身高。问题也变得很年轻，但为此，更好地获得所有客户ID，离开加入年龄和高度~~

最佳答案是接受的答案，我把它留在这里因为我学习了MySQL COALESCE（）函数，XD

如何创建处理丢失记录的SQL查询？

4 个答案: