Question

我有以下查询：

select 
    count(*) as leads, 
    (select count(*) from assignments where lead_id=leads.id and deleted_at is null) as assignments, 
    (select count(*) from assignments where lead_id=leads.id and deleted_at is not null) as returns, 
    date_format(leads.updated_at, "%m/%d/%Y") as date 
from `leads` where leads.updated_at between "2017-08-24 04:00:00" and "2017-08-26 03:59:59"
group by `date`

这会产生以下错误：

SQLSTATE [42000]：语法错误或访问冲突：1055表达式＃2 SELECT列表不在GROUP BY子句中并包含nonaggregated 列'leadbind.leads.id'在功能上不依赖于 GROUP BY子句中的列;这是不相容的 sql_mode = only_full_group_by（SQL：选择count（）作为引导，（选择从分配中计算（），其中lead_id = leads.id和deleted_at是 null）作为赋值，（从赋值中选择count（*）） lead_id = leads.id和deleted_at不为null）作为返回， date_format（leads.updated_at，“％m /％d /％Y”）作为leads的日期 “2017-08-24 04:00:00”和“2017-08-26”之间的leads.updated_at 03:59:59“分组date）

是否可以在保持group by date的同时更正此问题？我想保留子查询。

Answer 1

错误是常见错误。它来自关于GROUP BY查询的规则：您不能让列产生模糊结果。 select-list中的所有列必须位于GROUP BY子句中，或者位于聚合函数内，或者在功能上依赖于GROUP BY子句中的列。

有关此规则的详细信息，请阅读https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html

您的子查询列returns和id违反了此规则。每date有多个id。因此，您希望计算结果的分配和返回数量是多少？我猜你想要在给定日期内所有select count(*) as leads, SUM((select count(*) from assignments where lead_id=leads.id and deleted_at is null)) as assignments, SUM((select count(*) from assignments where lead_id=leads.id and deleted_at is not null)) as returns, date_format(leads.updated_at, '%m/%d/%Y') as date from `leads` where leads.updated_at between '2017-08-24 04:00:00' and '2017-08-26 03:59:59' group by `date`值的总分配和回报。

select 
    count(*) as leads, 
    COUNT(CASE WHEN a.deleted_at is null THEN a.lead_id END)) as assignments, 
    COUNT(CASE WHEN a.deleted_at is not null THEN a.lead_id END)) as returns,
    date_format(leads.updated_at, '%m/%d/%Y') as date 
from `leads` as l
left outer join `assignments` as a on l.id = a.lead_id
where leads.updated_at between '2017-08-24 04:00:00' and '2017-08-26 03:59:59'
group by `date`

我将字符串分隔符更改为更加标准的单引号。

我会进一步使用连接编写此查询，而不是执行两个相关的子查询：

COUNT(expr)

{{1}}不计算 expr 为空的行。

Answer 2

应该是

group by date_format(leads.updated_at, "%m/%d/%Y") , leads.id

在执行sql层次结构

中选择之前，分组依次排在第一位

2 个答案: