MYSQL选择多个不同值的总和

我的桌子看起来像这样：

| id | Vendor | Issue     |
|----|--------|-----------|
| 1  | Acme   | Defective |
| 2  | Best   | Returned  |
| 3  | Ace    | Other     |
| 4  | Best   | Returned  |
| 5  | Acme   | Other     |
| 6  | Ace    | Other     |
| 7  | Best   | Defective |

我需要一份Select报表来汇总每个供应商的每个不同问题的金额

select语句的输出在表中如下所示：

| Vendor | Defective | Returned | Other |
|--------|-----------|----------|-------|
| Acme   | 1         | 0        | 1     |
| Best   | 1         | 2        | 0     |
| Ace    | 0         | 0        | 2     |

---

任何帮助都将不胜感激。

您可以使用

CASE

子句分隔金额，如下所示：

select
  vendor,
  sum(case when issue = 'Defective' then 1 end) as defective,
  sum(case when issue = 'Returned' then 1 end) as returned,
  sum(case when issue = 'Other' then 1 end) as other
from my_table
group by vendor

最后声明：

$sql = "select
vendor,
sum(case when issue = 'Item Defective' THEN 1 ELSE 0 END) as 'defective',
sum(case when issue = 'Incorrect Item Received' THEN 1 ELSE 0 END) as 'received',
sum(case when issue = 'Incorrect Item Ordered' THEN 1 ELSE 0 END) as 'ordered',
sum(case when issue = 'Item Not Made to Drawing' THEN 1 ELSE 0 END) as 'drawing',
sum(case when issue = 'Other' THEN 1 ELSE 0 END) as 'other'
FROM record GROUP BY vendor";

---

谢谢你的帮助。它不起作用。尝试此操作：

$sql=“选择供应商，将sum（问题时案例='Item Defective'然后1 ELSE 0 END）作为'Defective'，将sum（问题时案例='error Item Received'，然后1 ELSE 0 END）作为'Received'，将sum（问题时案例='error Item Ordered'，然后1 ELSE 0 END）作为'Ordered'，sum（问题时的案例='Item Not Make to Drawing'然后1 ELSE 0 END）为'Drawing'，按供应商从记录组中求和（问题时的案例='Other'然后1 ELSE 0 END）为'Other'；但这也不起作用。没有错误-什么也不返回。

是否引发错误（哪一个），它是否给了您错误的结果？另外，您不需要

案例中的
ELSE
。它默认为
null
。谢谢！在修复之前删除。现在可以工作了。