#1055 - SELECT 列表的表达式 #6 不在 GROUP BY 子句中，并且包含非聚合列

Question

我在 MySQL 5.7 中对下面的查询有疑问，但在 MySQL 5.6 中它运行良好。

此消息每次都会出现：

1055 - SELECT 列表的表达式 #6 不在 GROUP BY 子句中，并且包含非聚合列
“electricity_databases.electricity_invoices.date_inserted”在功能上不依赖于 GROUP BY 子句中的列；这与 sql_mode=only_full_group_by 不兼容

SQL 代码：

SELECT 
homes.id,
homes.homeName,
homes.city, 
homes.date_registered,
ROUND(SUM(electricity_invoices.total), 2) AS TotalPrice,
DATEDIFF(NOW(), electricity_invoices.date_inserted) AS last_insert_in_days,
MAX(electricity_invoices.date_inserted) AS last_insert,
COUNT(electricity_invoices.homeID) AS countPaymentTimes,
MAX(electricity_invoices.currRead) AS currRead,
MAX(electricity_invoices.prevRead) AS prevRead,
ROUND(MAX(electricity_invoices.currRead) - MAX(electricity_invoices.prevRead), 1) AS lastComp,
customer.name

FROM homes

LEFT JOIN electricity_invoices ON
homes.id = electricity_invoices.homeID

LEFT JOIN customer ON
homes.id = customer.homeID

GROUP BY homes.id
ORDER BY homes.id

Answer 1

原因是在最新版本的 MySQL 中，默认情况下不允许在 group by 子句中添加非聚合列。您可以通过禁用 sql_mode from full group by 模式来禁用此行为。

只需在group by 子句中添加非聚合列即可。

select homes.id,
    homes.homeName,
    homes.city,
    homes.date_registered,
    ROUND(SUM(electricity_invoices.total), 2) as TotalPrice,
    DATEDIFF(NOW(), electricity_invoices.date_inserted) as last_insert_in_days,
    MAX(electricity_invoices.date_inserted) as last_insert,
    COUNT(electricity_invoices.homeID) as countPaymentTimes,
    MAX(electricity_invoices.currRead) as currRead,
    MAX(electricity_invoices.prevRead) as prevRead,
    ROUND(MAX(electricity_invoices.currRead) - MAX(electricity_invoices.prevRead), 1) as lastComp,
    customer.name
from homes
left join electricity_invoices on homes.id = electricity_invoices.homeID
left join customer on homes.id = customer.homeID
group by homes.id,
    homes.homeName,
    homes.city,
    homes.date_registered,
    customer.name

Answer 2

从 mysql 5.7 开始，您可以选择组中不存在的非聚合列 如果您想要与以前版本相同的行为，您必须撤销 sql_mode=only_full_group_by （使用 SET sql_mode = ''） 或更恰当地说，您应该建立一个选择或评估以按所有未聚合的列进行分组 例如：

SELECT 
homes.id,
homes.homeName,
homes.city, 
homes.date_registered,
ROUND(SUM(electricity_invoices.total), 2) AS TotalPrice,
DATEDIFF(NOW(), electricity_invoices.date_inserted) AS last_insert_in_days,
MAX(electricity_invoices.date_inserted) AS last_insert,
COUNT(electricity_invoices.homeID) AS countPaymentTimes,
MAX(electricity_invoices.currRead) AS currRead,
MAX(electricity_invoices.prevRead) AS prevRead,
ROUND(MAX(electricity_invoices.currRead) - MAX(electricity_invoices.prevRead), 1) AS lastComp,
customer.name

FROM homes

LEFT JOIN electricity_invoices ON
homes.id = electricity_invoices.homeID

LEFT JOIN customer ON
homes.id = customer.homeID

GROUP BY homes.id, homes.homeName,homes.city, homes.date_registered, DATEDIFF(NOW(), electricity_invoices.date_inserted) AS last_insert_in_days
ORDER BY homes.id

或者，由于您不需要这些列的特定值，因此对这些列使用（假）聚合

SELECT 
homes.id,
min(homes.homeName),
min(homes.city), 
min(homes.date_registered),
ROUND(SUM(electricity_invoices.total), 2) AS TotalPrice,
min(DATEDIFF(NOW(), electricity_invoices.date_inserted) AS last_insert_in_days),
MAX(electricity_invoices.date_inserted) AS last_insert,
COUNT(electricity_invoices.homeID) AS countPaymentTimes,
MAX(electricity_invoices.currRead) AS currRead,
MAX(electricity_invoices.prevRead) AS prevRead,
ROUND(MAX(electricity_invoices.currRead) - MAX(electricity_invoices.prevRead), 1) AS lastComp,
customer.name

FROM homes

LEFT JOIN electricity_invoices ON
homes.id = electricity_invoices.homeID

LEFT JOIN customer ON
homes.id = customer.homeID

GROUP BY homes.id
ORDER BY homes.id