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explain select id, nome from bea_clientes where id in (
     select group_concat(distinct(bea_clientes_id)) as list
     from bea_agenda
     where bea_clientes_id>0
     and bea_agente_id in(300006,300007,300008,300009,300010,300011,300012,300013,300014,300018,300019,300020,300021,300022)
)

当我尝试执行上述操作时（没有解释），MySQL 只是忙于使用 DEPENDENT SUBQUERY，这使得这变得非常缓慢。问题是优化器为客户端中的每个 id 计算子查询的原因。我什至将 IN 参数放在 group_concat 中，相信将结果作为纯“字符串”放置以避免扫描是相同的。

我认为这对于 5.5+ 的 MySQL 服务器不会有问题？MariaDb 中的测试也是如此。

这是一个已知的错误？我知道我可以将其重写为连接，但这仍然很糟糕。

Generated by: phpMyAdmin 4.4.14 / MySQL 5.6.26
Comando SQL: explain select id, nome from bea_clientes where id in ( select group_concat(distinct(bea_clientes_id)) as list from bea_agenda where bea_clientes_id>0 and bea_agente_id in(300006,300007,300008,300009,300010,300011,300012,300013,300014,300018,300019,300020,300021,300022) );
Lines: 2

 Current selection does not contain a unique column. Grid edit, checkbox, Edit, Copy and Delete features are not available.

| id | select_type        | table        | type  | possible_keys                 | key           | key_len | ref  | rows  | Extra                              |
|----|--------------------|--------------|-------|-------------------------------|---------------|---------|------|-------|------------------------------------|
| 1  | PRIMARY            | bea_clientes | ALL   | NULL                          | NULL          | NULL    | NULL | 30432 | Using where                        |
| 2  | DEPENDENT SUBQUERY | bea_agenda   | range | bea_clientes_id,bea_agente_id | bea_agente_id | 5       | NULL | 2352  | Using index condition; Using where |
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1 回答 1

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显然很难在没有数据的情况下进行测试，但如下所示。子查询在 mysql 中并不好（尽管它是我的首选引擎）。我还可以建议对相关列进行索引，这将提高两个查询的性能。为了清楚起见，我还可以建议扩展查询。

select t1.id,t1.nome from (
    (select group_concat(distinct(bea_clientes_id)) as list from bea_agenda where bea_clientes_id>0 and bea_agente_id in                    (300006,300007,300008,300009,300010,300011,300012,300013,300014,300018,300019,300020,300021,300022)
    ) as t1
    join
    (select id, nome from bea_clientes) as t2   
    on t1.list=t2.id
)
于 2016-04-27T17:56:33.407 回答