sqlfiddle.com 5.5.30和MariaDB 5.5.31中的结果不同

时间:2013-07-09 11:29:38

标签: mysql mariadb sqlfiddle

sqlfiddle:http://sqlfiddle.com/#!2/9a8b3/1

从小提琴中获取结构和数据以及查询, 导入我的MariaDB 5.5.31,我得到不同的结果:

sqlfiddle

PID  NAME       LEAGUEPOINTS        TOTALLEAGUEPOINTS
2   Peter   16,13,9,4,2            44
1   Daniel  3425,543,234,43,29,22,21,21,19,17,13,12,12,12,11,9,9,9,8,7      4476

MariaDB的

pid  name    leaguepoints       totalleaguepoints   
2   Peter   16,13,9,4,2             44
1   Daniel  3425,543,234,43,29,22,21,21,19,17,13,12,12,12,11,9,9,9,8,7,7,6,5,5,4,4,4,3,3,2,1    4520

查询:

SELECT                
    p.pid,
    p.name,   
    GROUP_CONCAT( gC.leaguepoints ORDER BY leaguepoints DESC ) AS leaguepoints, 
    SUM(gC.leaguepoints) AS totalleaguepoints
FROM test_golf_player p
LEFT JOIN 
(
    SELECT pid, leaguepoints, @Sequence:=IF(@PrevPid = pid, @Sequence + 1, 0) AS aSequence, @PrevPid := pid
    FROM
    (
        SELECT pid, leaguepoints
        FROM test_golf_card 
        ORDER BY pid, leaguepoints DESC
    ) Sub1
    CROSS JOIN (SELECT @PrevPid := 0, @Sequence := 0) Sub2
) gC
ON p.pid = gC.pid AND aSequence < 20
GROUP BY p.pid
ORDER BY p.name DESC 

知道为什么吗?

1 个答案:

答案 0 :(得分:1)

害怕我没有手中的MariaDB,但是你可以尝试以下方法来看看如何输出用户变量: -

SELECT  *
FROM test_golf_player p
LEFT JOIN 
(
    SELECT pid, leaguepoints, @Sequence:=IF(@PrevPid = pid, @Sequence + 1, 0) AS aSequence, @PrevPid := pid
    FROM
    (
        SELECT pid, leaguepoints
        FROM test_golf_card 
        ORDER BY pid, leaguepoints DESC
    ) Sub1
    CROSS JOIN (SELECT @PrevPid := 0, @Sequence := 0) Sub2
) gC
ON p.pid = gC.pid 
ORDER BY p.name DESC 

编辑 - 在查看结果时进行一些调查似乎MariaDB忽略了子查询中的ORDER BY。因此,序列号是随机顺序,并且当pid改变时也会重置(由于顺序不固定,它会随机重复)。有点谷歌,似乎这是MariaDB的故意特征。 SQL标准将表定义为无序行集,子选择被视为表,因此忽略了order by - https://kb.askmonty.org/en/why-is-order-by-in-a-from-subquery-ignored/

这有点不利。不确定是否有解决办法,因为我现在想不到一个。对于这个要处理的原始问题,我认为有必要使用可能效率不高的相关子选择。