如何制作处理缺失记录的SQL查询？

Question

我有两个表，其中包含客户的年龄和高度。

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 | 
+-----------+------------+ 

我需要编写一个可读取所有年龄和高度的SELECT查询。因此，像这样......

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

然而（和这里的扭曲）由于马虎记录保存，有两个表丢失记录。 （例如，年龄中的顾客ID 5和高度上的顾客ID 3）。

有没有办法编写查询，以便它仍然可以工作，但每当数据丢失时返回零？

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

一条路可走（他们是别人，总是）

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; 

看到SqlFiddle

Answer 2

使用LEFT JOIN与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; 

OK OK，于是赶紧回答... ...以上才会给你0作为高度。问题是得到同样的年龄，但对于这一点，最好把所有客户的ID，左连接的年龄和高度

最好的答案是公认的答案，我离开这里，是因为我学到了MySQL的COALESCE（）函数，XD

Answer 3

MySQL没有全外连接，但你可以simulate one与LEFT JOIN，随后RIGHT JOIN，具有UNION相结合，将结合+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 4

你真正需要的是一个full outer join，但MySQL不支持这一点。相反，让所有的客户在子查询和使用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;