Question

如何在以下方案中计算每hit_type个唯一（click = campaign）的数量？：

CREATE TABLE statistics
    (`id` int, `uid` int, `hit_type` varchar(10), `name` varchar(55), `date` date)
;

INSERT INTO statistics
    (`id`, `uid`, `hit_type`, `name`, `date`)
VALUES
    (1,'100','visit','campaign1','2015-03-18'),
    (2,'100','visit','campaign1','2015-03-19'),
    (3,'100','click','campaign1','2015-03-18'),
    (4,'100','click','campaign1','2015-03-18'),
    (5,'100','click','campaign1','2015-03-20'),
    (6,'100','client','campaign1','2015-03-19'),
    (7,'100','client','campaign1','2015-03-20'),

    (8,'200','visit','campaign1','2015-03-19'),
    (9,'200','visit','campaign1','2015-03-19'),
    (10,'200','visit','campaign2','2015-03-20'),
    (11,'200','click','campaign1','2015-03-18'),
    (12,'200','click','campaign1','2015-03-19'),
    (13,'200','client','campaign1','2015-03-19'),
    (14,'200','client','campaign2','2015-03-20')

;
SELECT `name`, 
    SUM(IF(`hit_type` = 'click',1,0)) as click , 
    SUM(IF(`hit_type` = 'visit',1,0)) as visit, 
    SUM(IF(`hit_type` = 'client',1,0)) as client 
FROM `statistics` 
WHERE `name` IN('campaign1','campaign2') 
GROUP BY `name` 
ORDER BY `name`

Answer 1

如果您正在寻找独特的日期，那么您可以：

SELECT `name`, 
        COUNT(DISTINCT CASE WHEN `hit_type` = 'click' THEN date END) as click , 
        COUNT(DISTINCT CASE WHEN `hit_type` = 'visit' THEN date END) as visit, 
        COUNT(DISTINCT CASE WHEN `hit_type` = 'client' THEN date END) as client 
FROM `statistics` 
WHERE `name` IN('campaign1','campaign2') 
GROUP BY `name` 
ORDER BY `name`;

SQL SUM不同的值

1 个答案: