Question

您好我有这样的SQL查询：

SELECT 
    COUNT(*) AS total, 
    COUNT(referrer) AS referrer,  
    DATE_FORMAT(created_at, "%Y") AS YEAR, 
    DATE_FORMAT(created_at, "%m") AS MONTH, 
    DATE_FORMAT(created_at, "%d") AS DAY
FROM 
    users
GROUP BY 
    EXTRACT(DAY FROM created_at), 
    EXTRACT(MONTH FROM created_at), 
    EXTRACT(YEAR FROM created_at)
ORDER BY 
    created_at ASC

这为我提供了按日（月和年）排序的总表。我需要的是添加一个子查询来显示一个具有条件的count（）。类似的东西：

(SELECT count(*) FROM users WHERE result = '1') as winners

问题显然在于主查询的分组，因为我得到相同的结果而没有为每一行分组子查询。

执行此类查询的正确方法是什么？

Answer 1

怎么样

SELECT 
    COUNT(*) AS total, 
    COUNT(referrer) AS referrer,  
    SUM(case when result = '1' then 1 else 0 end) as winners
    DATE_FORMAT(created_at, "%Y") AS YEAR, 
    DATE_FORMAT(created_at, "%m") AS MONTH, 
    DATE_FORMAT(created_at, "%d") AS DAY
FROM 
    users
GROUP BY 
    EXTRACT(DAY FROM created_at), 
    EXTRACT(MONTH FROM created_at), 
    EXTRACT(YEAR FROM created_at)
ORDER BY 
    created_at ASC;

假设created_at是一个时间戳，我们可以重新编写您的查询，只需要一个字段进行分组，如下所示：

SELECT 
    COUNT(*) AS total, 
    COUNT(referrer) AS referrer,  
    SUM(case when result = '1' then 1 else 0 end) as winners
    DATE_FORMAT(created_at, "%Y") AS YEAR, 
    DATE_FORMAT(created_at, "%m") AS MONTH, 
    DATE_FORMAT(created_at, "%d") AS DAY
FROM 
    users
GROUP BY date(created_at)
ORDER BY 
    created_at ASC;

请注意我手边没有mysql。

Answer 2

您可以使用UNION

加入这两个查询

SELECT 
    COUNT(*) AS total, 
    COUNT(referrer) AS referrer,  
    DATE_FORMAT(created_at, "%Y") AS YEAR, 
    DATE_FORMAT(created_at, "%m") AS MONTH, 
    DATE_FORMAT(created_at, "%d") AS DAY
FROM 
    users
GROUP BY 
    EXTRACT(DAY FROM created_at), 
    EXTRACT(MONTH FROM created_at), 
    EXTRACT(YEAR FROM created_at)
ORDER BY 
    created_at ASC
UNION
SELECT count(*) FROM users 
GROUP BY 
    EXTRACT(DAY FROM created_at), 
    EXTRACT(MONTH FROM created_at), 
    EXTRACT(YEAR FROM created_at)
HAVING result = '1'

如何在子查询中保留分组

2 个答案: