SQLSTATE[42000]: Syntax error or access violation: 1055 'kf_local.g.group_name' isn't in GROUP BY

这是 ThinPHP 中的一段取数据的代码：

$result = $this->field('g.group_code,g.group_name,count(m.member_code) AS member_num,c.company_name,u.user_uid')
                       ->alias('g')
                       ->join('LEFT JOIN company AS c ON g.company_code=c.company_code LEFT JOIN user AS u ON u.user_code=g.holder_code LEFT JOIN groups_member AS m ON m.group_code=g.group_code')
                       ->where("g.group_name LIKE '%" . $key . "%' OR u.user_uid='" . $key ."'")
                       ->group('g.group_code,g.group_name,c.company_name,u.user_uid')
                       ->select();

生成如下 SQL :

SELECT 
  g.group_code,
  g.group_name,
  COUNT(m.member_code) AS member_num,
  c.company_name,
  u.user_uid 
FROM
  groups g 
  LEFT JOIN company AS c 
    ON g.company_code = c.company_code 
  LEFT JOIN USER AS u 
    ON u.user_code = g.holder_code 
  LEFT JOIN groups_member AS m 
    ON m.group_code = g.group_code 
WHERE (
    g.group_name LIKE '%武汉%' 
    OR u.user_uid = '武汉'
  ) 
GROUP BY g.group_code

乍一看是没什么问题，直接丢到数据库执行也是没什么问题。

“但是” 程序运行就报错，我们来看看错误提示：

SQLSTATE[42000]: Syntax error or access violation: 1055 'kf_local.g.group_name' isn't in GROUP BY

这是 Overflow 里一段与之相关的答案，

You need to have a full group by:

SELECT name, type, language, code
FROM users
WHERE verified = ‘1’
GROUP BY name, type, language, code
ORDER BY count DESC LIMIT 0, 25

SQL92 requires that all columns (except aggregates) in the select clause is part of the group by clause.

create table t (x int, y int);
insert into t (x,y) values (1,1),(1,2),(1,3);
select x,y from t group by x;

+——+——+
| x | y |
+——+——+
| 1 | 1 |
+——+——+

I.e. a random y is select for the group x. One can prevent this behavior by setting > @@sql_mode:
set @@sql_mode=’ONLY_FULL_GROUP_BY’;
select x,y from t group by x;
ERROR 1055 (42000): ‘test.t.y’ isn’t in GROUP BY

其实也就是 sql_mode 的问题，可以试试改为默认的 “宽松”模式：

set @@sql_mode='';


SELECT 
  g.group_code,
  g.group_name,
  COUNT(m.member_code) AS member_num,
  c.company_name,
  u.user_uid 
FROM
  groups g 
  LEFT JOIN company AS c 
    ON g.company_code = c.company_code 
  LEFT JOIN USER AS u 
    ON u.user_code = g.holder_code 
  LEFT JOIN groups_member AS m 
    ON m.group_code = g.group_code 
WHERE (
    g.group_name LIKE '%武汉%' 
    OR u.user_uid = '武汉'
  ) 
GROUP BY u.user_uid

转载于:https://www.cnblogs.com/peterzha/p/6853249.html