没有组聚合列由

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` 

这产生了以下错误：

SQLSTATE[42000]: Syntax error or access violation: 1055 Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'leadbind.leads.id' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by (SQL: 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)

是否有可能解决此同时仍保持group by date？我想保留子查询。

Answer 1

应该

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

集团首先谈到选择之前执行

Answer 2

的错误是常见的一种SQL的层次。它来自有关GROUP BY查询的规则：您不能让列产生不明确的结果。您的选择列表中的所有列必须位于GROUP BY子句中，或位于聚合函数内，否则在功能上依赖于GROUP BY子句中的列。

，详细了解这个规则，阅读https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html

你的子查询列assignments和returns违反了这一规则。每个date有多个id。因此，您希望计算结果的数量是多少？我想你会希望给定日期的所有id值的总分配和返回。

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` 

我将字符串分隔符更改为更为标准的单引号。

我将使用一个连接，而不是做两分相关子查询进一步编写这个查询：

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` 

COUNT(expr)其中EXPR为空不计行。