Question

我想计算一下，如果promotion_id是1483或1887，confirm_at不为null的次数。

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

示例：

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

所以，结果将是：

status_inv  not_nul_specific_promotions
before      2
after       3

PS。我想在select语句中这样做。我不想添加“where”。

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

将IN测试放在WHERE子句中，以便只计算这些行：

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

您也不需要IF，因为在MySQL布尔值中，TRUE和FALSE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

您也可以使用COUNT(confirmed_at)代替SUM()，因为COUNT()仅计算非空值。

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

或者将两个测试都放在WHERE：

中

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

如果您不想使用WHERE，可以在AND测试中使用IF：

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，这是查询：

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

mysql：如果求和则嵌套

2 个答案: