为什么SELECT结果在mysql和sqlite之间有所不同? [英] Why does SELECT results differ between mysql and sqlite?

问题描述

我在问这个问题以简化和扩展的方式.

考虑以下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       

并且mysql返回:

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

据我所知，mysql的结果是正确的，而sqlite的结果是不正确的.我尝试使用sqlite强制转换为real，如下所示，但它仍返回两条记录:

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返回两个记录?

快速更新:

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

另一个更新:

我已将有关此问题的电子邮件发送至[e-mail].

我自己，我一直在玩VDBE，发现了一些有趣的东西.我拆分了not exists的每个循环的执行跟踪(每个avg组一个).

要具有三个平均组，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应该变成了i:5:

我现在正试图找出原因.

最终编辑:

因此，我一直在充分利用sqlite源代码.我现在对野兽有了更好的了解，尽管我会让原始开发人员像他看来的那样进行整理已经做到了:

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

至少对我而言，有趣的是，似乎较新的版本(在我使用的版本之后多次)支持插入上述提交中添加的测试用例中使用的多条记录:

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

解决方案

我试图弄乱查询的某些变体.

似乎sqlite在嵌套的 HAVING 表达式中使用先前声明的字段时出错.

在您的示例中，avg1下的秒数始终等于5.0

看:

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);

此记录不返回任何内容，但是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票证列表中找不到任何类似的错误.. >

I'm re-asking this question in a simplified and expanded manner.

Consider these sql statements:

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);

Using sqlite, the select statement returns:

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

and mysql returns:

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

As far as I can tell, mysql's results are correct, and sqlite's are incorrect. I tried to cast to real with sqlite as in the following but it returns two records still:

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);

Why does sqlite return two records?

Quick update:

I ran the statement against the latest sqlite version (3.7.11) and still get two records.

Another update:

I sent an email to [e-mail] about the issue.

Myself, I've been playing with VDBE and found something interesting. I split the execution trace of each loop of not exists (one for each avg group).

To have three avg groups, I used the following statements:

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);

We clearly see that somehow what should be r:4.5 has become i:5:

I'm now trying to see why that is.

Final edit:

So I've been playing enough with the sqlite source code. I understand the beast much better now, although I'll let the original developer sort it out as he seems to already be doing it:

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

Interestingly, to me at least, it seems that the newer versions (some times after the version I'm using) supports inserting multiple records as used in a test case added in the aforementioned commit:

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

解决方案

I tried to mess with some variants of query.

It seems, like sqlite has errors in using of previous declared fields in a nested HAVING expressions.

In your example avg1 under second having is always equal to 5.0

Look:

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);

This one returns nothing, but execution of the following query returns both records:

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

I can not find any similar bug at sqlite tickets list.

这篇关于为什么SELECT结果在mysql和sqlite之间有所不同?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！