将两个类似的查询结合在一起

Question

我有两个独立工作的MySQL查询，我希望将它们组合在一起，以便返回三个值。

查询1个检查多少帐户已被删除：

SELECT 
    COUNT(1) AS deleted_count, 
    SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date 
FROM 
    exit_reasons e 
WHERE 
    e.timestamp>='$sixmonths' 
GROUP BY 
    WEEKOFYEAR(e.timestamp) 
ORDER BY 
    display_date ASC 
LIMIT 26 

这返回的日期，谁在这一周中删除

查询2个检查多少这些都随后再次报名人数：

SELECT 
    COUNT(1) AS date_count, 
    SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date 
FROM 
    exit_reasons e 
LEFT JOIN 
    companies c on e.email=c.email 
WHERE 
    e.timestamp>='$sixmonths' AND c.email IS NOT NULL 
GROUP BY 
    WEEKOFYEAR(e.timestamp) 
ORDER BY 
    display_date ASC 
LIMIT 26 

这会返回一个日期和星期删除的号码谁现在有一个新的帐户

我想它返回一个日期，然后数删除，并在一个查询再结合数，所以我尝试：

SELECT 
    COUNT(1) AS date_count, 
    SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date, 
    date_count as rejoined_count from 
     (SELECT 
      COUNT(1) AS date_count, 
      SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date 
     FROM 
      exit_reasons e2 
     LEFT JOIN 
      companies c on e.email=c.email 
     LEFT JOIN 
      companies_users cu on e.email=cu.email 
     WHERE 
      e2.timestamp>='$sixmonths' AND c.email IS NOT NULL 
     GROUP BY 
      WEEKOFYEAR(e.timestamp) 
     ORDER BY 
      display_date ASC 
     LIMIT 26) 
FROM 
    exit_reasons e 
WHERE 
    e.timestamp>='$sixmonths' 
GROUP BY 
    WEEKOFYEAR(e.timestamp) 
ORDER BY 
    display_date ASC 
LIMIT 26 

，但我得到一个语法错误 - 我怎么可以将这些查询组合在一起成为一个查询？

Answer 1

通过使用聚合函数以及一些条件逻辑（如CASE表达式），您应该能够将两个查询合并为一个查询：

SELECT 
    COUNT(1) AS deleted_count, 
    SUM(CASE WHEN c.email IS NOT NULL THEN 1 ELSE 0 END) as date_count, 
    SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date 
FROM exit_reasons e 
LEFT JOIN companies c 
    on e.email=c.email 
WHERE e.timestamp>='$sixmonths' 
GROUP BY WEEKOFYEAR(e.timestamp) 
ORDER BY display_date ASC 
LIMIT 26; 

See Demo。如果将c.email IS NOT NULL移动到SUM(CASE..中，则可以检查第二个查询，该查询允许您获取总数不为空的行。

Answer 2

我认为下面会做你想要什么：

SELECT COUNT(*) AS deleted_count, 
     COUNT(c.email) as date_count, 
     SUBDATE(e.timestamp, INTERVAL WEEKDAY(e.timestamp) DAY) AS display_date 
FROM exit_reasons e LEFT JOIN 
    companies c 
    on e.email = c.email 
WHERE e.timestamp >= '$sixmonths' 
GROUP BY WEEKOFYEAR(e.timestamp) 
ORDER BY display_date ASC 
LIMIT 26; 

倘若有人可以注册多个具有相同电子邮件一次，你应该改变count()使用distinct：

COUNT(DISTINCT e.email) as deleted_count, 
COUNT(DISTINCT c.email) as date_count