MySQL的：如果嵌套总结

Question

我想指望有多少次confirmed_at没有空给一个promotion_id或者是1483或1887年。

select 
    IF(promotion_id IN(1483,1887), sum(IF(confirmed_at IS NOT NULL,1,0)),0) as all_us_vs_voda, 
    status_inv 
from blabla 

group by status_inv 

例子：

promotion_id status_inv confirmed_at 
1   'before' 2016-05 
1483   'before' NULL 
1483   'after'  2016-05 
1483   'after'  2016-07 
1887   'before' 2016-08 
1887   'before' 2017-09 
1887   'after' 2017-09 

因此，其结果必然成为：

status_inv not_nul_specific_promotions 
before  2 
after  3 

PS。我想在select语句中做到这一点。我不想添加“哪里”。

事实上，我能够这样做：

sum(IF(promotion_id IN(1483,1887), IF(confirmed_at IS NOT NULL,1,0),0)) as all_us_vs_voda, 

Answer 1

把IN测试WHERE子句中，所以你只算那些行：

SELECT status_inv, SUM(IF(confirmed_at IS NOT NULL, 1, 0) AS not_nul_specific_promotions 
FROM blabla 
WHERE promotion_id IN (1483, 1887) 
GROUP BY status_inv 

你也不需要IF，因为在MySQL中，对于TRUE和FALSE，布尔值评估为1和0。

SELECT status_inv, SUM(confirmed_at IS NOT NULL) AS not_nul_specific_promotions 
FROM blabla 
WHERE promotion_id IN (1483, 1887) 
GROUP BY status_inv 

你也可以只使用COUNT(confirmed_at)代替SUM()，因为COUNT()只统计非空值。

SELECT status_inv, COUNT(confirmed_at) AS not_nul_specific_promotions 
FROM blabla 
WHERE promotion_id IN (1483, 1887) 
GROUP BY status_inv 

或者把两个试验中WHERE：

SELECT status_inv, COUNT(*) AS not_nul_specific_promotions 
FROM blabla 
WHERE promotion_id IN (1483, 1887) AND confirmed_at IS NOT NULL 
GROUP BY status_inv 

如果你不想使用WHERE，你可以在IF测试使用AND：

SELECT status_inv, SUM(IF(confirmed_at IS NOT NULL AND promotion_id IN (1483, 1887), 1, 0) AS not_nul_specific_promotions 
FROM blabla 
GROUP BY status_inv 

Answer 2

您可以使用count为此，这里是查询：

select STATUS_INV,count(case when PROMOTION_ID=1483 THEN 1 when PROMOTION_ID =1887 THEN 1 ELSE NULL END) 
as confirmed_at FROM sample WHERE confirmed_at IS NOT NULL group by STATUS_INV