Question

我以简化和扩展的方式重新询问question。

考虑这些sql语句：

create table foo (id INT, score INT);

insert into foo values (106, 4);
insert into foo values (107, 3);
insert into foo values (106, 5);
insert into foo values (107, 5);

select T1.id, avg(T1.score) avg1
from foo T1
group by T1.id
having not exists (
    select T2.id, avg(T2.score) avg2
    from foo T2
    group by T2.id
    having avg2 > avg1);

使用sqlite，select语句返回：

id          avg1      
----------  ----------
106         4.5       
107         4.0

和mysql返回：

+------+--------+
| id   | avg1   |
+------+--------+
|  106 | 4.5000 |
+------+--------+

据我所知，mysql的结果是正确的，sqlite是不正确的。我尝试使用sqlite转换为real，如下所示，但它仍然返回两个记录：

select T1.id, cast(avg(cast(T1.score as real)) as real) avg1
from foo T1
group by T1.id
having not exists (
    select T2.id, cast(avg(cast(T2.score as real)) as real) avg2
    from foo T2
    group by T2.id
    having avg2 > avg1);

为什么sqlite会返回两条记录？

快速更新：

我针对最新的sqlite版本（3.7.11）运行了该语句，仍然获得了两条记录。

另一次更新：

我发送了一封关于此问题的电子邮件至sqlite-users@sqlite.org。

我自己，我一直在玩VDBE，发现了一些有趣的东西。我将每个not exists循环的执行跟踪分开（每个平均组一个）。

为了拥有三个平均组，我使用了以下语句：

create table foo (id VARCHAR(1), score INT);

insert into foo values ('c', 1.5);
insert into foo values ('b', 5.0);
insert into foo values ('a', 4.0);
insert into foo values ('a', 5.0);

PRAGMA vdbe_listing = 1;
PRAGMA vdbe_trace=ON;

select avg(score) avg1
from foo
group by id
having not exists (
    select avg(T2.score) avg2
    from foo T2
    group by T2.id
    having avg2 > avg1);

我们清楚地看到r:4.5应该变成i:5：

enter image description here

我现在正试图了解原因。

最终修改：

所以我一直在玩sqlite源代码。我现在对这种野兽的理解要好得多，虽然我会让original developer对它进行排序，因为他似乎已经在做了这件事：

http://www.sqlite.org/src/info/430bb59d79

有趣的是，至少对我来说，似乎新版本（在我使用的版本之后的某些时候）支持插入在上述提交中添加的测试用例中使用的多个记录：

CREATE TABLE t34(x,y);
INSERT INTO t34 VALUES(106,4), (107,3), (106,5), (107,5);

Answer 1

我试图弄乱一些查询变种。

似乎sqlite在嵌套 HAVING 表达式中使用先前声明的字段时出错。

在你的例子avg1下，第二个总是等于5.0

查找

select T1.id, avg(T1.score) avg1
from foo T1
group by T1.id
having not exists (
    SELECT 1 AS col1 GROUP BY col1 HAVING avg1 = 5.0);

这个没有返回任何内容，但执行以下查询将返回两个记录：

...
having not exists (
    SELECT 1 AS col1 GROUP BY col1 HAVING avg1 <> 5.0);

我在sqlite tickets list找不到任何类似的错误。

Answer 2

让我们看看这两种方式，我将使用postgres 9.0作为我的参考数据库

（1）

-- select rows from foo 

select T1.id, avg(T1.score) avg1
from foo T1
group by T1.id
-- where we don't have any rows from T2
having  not exists (
-- select rows from foo
select T2.id, avg(T2.score) avg2
from foo T2
group by T2.id
-- where the average score for any row is greater than the average for 
-- any row in T1
having avg2 > avg1);

 id  |        avg1        
-----+--------------------
 106 | 4.5000000000000000
(1 row)

然后让我们移动子查询中的一些逻辑，摆脱'not'：（2）

-- select rows from foo 
select T1.id, avg(T1.score) avg1
from foo T1
group by T1.id
-- where we do have rows from T2
having  exists (
-- select rows from foo
select T2.id, avg(T2.score) avg2
from foo T2
group by T2.id
-- where the average score is less than or equal than the average for any row in T1
having avg2 <= avg1);
-- I think this expression will be true for all rows as we are in effect doing a
--cartesian join 
-- with the 'having' only we don't display the cartesian row set

 id  |        avg1        
-----+--------------------
 106 | 4.5000000000000000
 107 | 4.0000000000000000
(2 rows)

所以你必须问问自己 - 当你在having子句中执行这个相关的子查询时，你实际上是什么意思，如果它根据主要查询的每一行评估每一行，我们正在进行笛卡尔加入，我不知道我们认为我们应该指责SQL引擎。

如果你想要每一行都小于最大平均值你应该说的是：

select T1.id, avg(T1.score) avg1 
from foo T1 group by T1.id
having avg1 not in 
(select max(avg1) from (select id,avg(score) avg1 from foo group by id))

Answer 3

你试过这个版本吗？：

select T1.id, avg(T1.score) avg1
from foo T1
group by T1.id
having not exists (
    select T2.id, avg(T2.score) avg2
    from foo T2
    group by T2.id
    having avg(T2.score) > avg(T1.score));

这一个（应该给出相同的结果）：

select T1.*
from
  ( select id, avg(score) avg1
    from foo 
    group by id
  ) T1
where not exists (
    select T2.id, avg(T2.score) avg2
    from foo T2
    group by T2.id
    having avg(T2.score) > avg1);

也可以使用派生表处理查询，而不是HAVING子句中的子查询：

select ta.id, ta.avg1
from 
  ( select id, avg(score) avg1
    from foo
    group by id
  ) ta
  JOIN
  ( select avg(score) avg1
    from foo 
    group by id
    order by avg1 DESC
    LIMIT 1
  ) tmp
  ON tmp.avg1 = ta.avg1

为什么mysql和sqlite之间的SELECT结果不同？

3 个答案: