MySQL与sql_mode = only_full_group_by不兼容

时间:2017-05-13 09:55:04

标签: mysql sql database mysql-error-1055

我有以下查询(更新见下文):

SELECT 

i0_.id AS id_0, 
i0_.address AS address_1, 
i1_.name AS name_2, 
c2_.name AS name_3, 
c3_.code AS code_4, 
l4_.iso AS iso_5, 
c5_.id AS id_6, 
c6_.name AS name_7, 
i7_.identifier AS identifier_8 

FROM institutions i0_ 
LEFT JOIN institution_languages i1_ ON (i1_.institution_id = i0_.id) 
LEFT JOIN countries c3_ ON (c3_.id = i0_.country_id) 
LEFT JOIN country_languages c2_ ON (c2_.country_id = c3_.id) 
LEFT JOIN country_spoken_languages c8_ ON (c8_.country_id = c3_.id) 
LEFT JOIN cities c5_ ON (c5_.id = i0_.city_id) 
LEFT JOIN city_languages c6_ ON (c6_.city_id = c5_.id) 
LEFT JOIN languages l4_ ON (l4_.id = i1_.language_id) 
LEFT JOIN institution_types i7_ ON (i0_.institution_type_id = i7_.id) 

WHERE c8_.is_primary = 1 
AND c8_.language_id = l4_.id 
AND c2_.language_id = 546 
AND i7_.identifier = "work_place" 
GROUP BY id_6 #here is the issue...
  

更新查询

SELECT 
i0_.id AS id_0, 
i0_.address AS address_1, 
i1_.name AS name_2, 
c2_.name AS name_3, 
c3_.code AS code_4, 
c4_.name AS name_5, 
i5_.identifier AS identifier_6 

FROM institutions i0_ 

LEFT JOIN institution_languages i1_ ON (i1_.institution_id = i0_.id) 
LEFT JOIN countries c3_ ON (c3_.id = i0_.country_id) 
LEFT JOIN country_languages c2_ ON (c2_.country_id = c3_.id AND c2_.language_id = ?) 
LEFT JOIN country_spoken_languages c6_ ON (c6_.country_id = c3_.id AND c6_.language_id = ? AND c6_.is_primary = ?) 
LEFT JOIN city_languages c4_ ON (c4_.city_id = i0_.city_id) 
LEFT JOIN institution_types i5_ ON (i0_.institution_type_id = i5_.id) 

WHERE i5_.identifier = ? 
GROUP BY i0_.city_id

此查询遇到GROUP_BY问题,我不知道如何解决:

  

1055 - SELECT列表的表达式#1不在GROUP BY子句中,并且包含非聚合列' database.i0_.id'这不是

     

在功能上依赖于GROUP BY子句中的列;这是   与sql_mode = only_full_group_by

不兼容

我知道这可以通过设置only_full_group_by轻松解决,但我可以对查询做些什么来使其正常工作而不必修改服务器上的MySQL设置?

2 个答案:

答案 0 :(得分:2)

如果您不想更改sql_mode = only_full_group_by

你可以简单地将一个aggegation函数添加到不参与group by的列中(例如min()或max()

(在previuos版本中,此列的结果是不可预测的。通过这种方式,您可以为获取这些列的值分配规则)

SELECT 
  i0_.id AS id_0, 
  min(i0_.address AS) address_1, 
  min(i1_.name) AS name_2, 
  min(c2_.name )AS name_3, 
  min(c3_.code) AS code_4, 
  min(c4_.name) AS name_5, 
  min(i5_.identifier) AS identifier_6 

FROM institutions i0_ 

LEFT JOIN institution_languages i1_ ON (i1_.institution_id = i0_.id) 
LEFT JOIN countries c3_ ON (c3_.id = i0_.country_id) 
LEFT JOIN country_languages c2_ ON (c2_.country_id = c3_.id AND c2_.language_id = ?) 
LEFT JOIN country_spoken_languages c6_ ON (c6_.country_id = c3_.id AND c6_.language_id = ? AND c6_.is_primary = ?) 
LEFT JOIN city_languages c4_ ON (c4_.city_id = i0_.city_id) 
LEFT JOIN institution_types i5_ ON (i0_.institution_type_id = i5_.id) 

WHERE i5_.identifier = ? 
GROUP BY i0_.city_id

答案 1 :(得分:0)

我认为你应该至少在select字段中使用聚合函数来使查询起作用,请参阅下面我尝试使用表'customers'和'orders'中的示例数据库'classicmodels':

select c.customerNumber, sum(o.orderNumber)
from customers c
join orders o
on(c.customerNumber = o.customerNumber)
group by c.customerNumber;

返回结果,但是如果我删除函数'sum()',则out为:

错误1055(42000):SELECT列表的表达式#2不在GROUP BY子句中,并且包含非聚合列'classicmodels.o.orderNumber',它在功能上不依赖于GROUP BY子句中的列;这与sql_mode = only_full_group_by

不兼容