mysql - 计算列中的不同值，如果在 Where 子句中找不到则返回零

Question

这是我的表：

我正在运行这个查询，

SELECT fruits,COUNT(*) as count FROM meva where owner = 'bill' GROUP BY fruits ORDER BY count 

我得到的结果是

如果与这样的所有者不匹配，是否可以将其他水果项目设为 0，

Fruit   | count
Apple   ,   2
Mango   ,   0
Banana  ,   0
Strawberry ,0
Appricot,   0
Alfonso ,   0

非常感谢对我的查询进行一些修改。谢谢

Answer 1

使用SUM()和CASE

SELECT  fruits, 
        SUM(CASE WHEN owner = 'bill' THEN 1 ELSE 0 END) as `count `
FROM     meva 
GROUP BY fruits 
ORDER BY fruits

或使用IF（仅在MySQL）

SELECT  fruits, 
        SUM(IF(owner = 'bill',1,0)) as `count `
FROM     meva 
GROUP BY fruits 
ORDER BY fruits

• SQLFiddle 演示（CASE 和 IF）

Answer 2

是的，上面有一个很好的答案。你也可以试试这个！

SELECT  fruits, 
        SUM(Coalesce(owner='bill',0)) as counts
FROM     meva 
GROUP BY fruits 
ORDER BY fruits
;