我有这样的预订表
place day name1 floor1 name2 floor2
--------- -------- --------- --------- --------- ---------
Anaheim Monday Sean 3rd Jenny 2nd
Anaheim Monday Jenny 2nd Sean 3rd
Cerritos Saturday Dennis 4th Sean 3rd
如您所见，第一行和第二行只是重复，顺序不同。 我想要一个像
这样的决赛桌place day name1 floor1 name2 floor2
--------- -------- --------- --------- --------- ---------
Anaheim Monday Sean 3rd Jenny 2nd
Cerritos Saturday Dennis 4th Sean 3rd
我在想这样的事情
SELECT t1.place, t1.day, t1.name1, t1.floor, t1.name2, t1.floor2
FROM table t1, table t2
WHERE NOT (t1.place = t2.place AND t1.day = t2.day AND
t1.name1 = t2.name2 AND t1.floor1 = t1.floor2);
但这不起作用:(也许我的理解太短了......
我该如何处理？
修改：
好吧，我应该提到这一点，但我正在寻找一个总体思路，而不是这个特定的表格。
可能有像
这样的表格place day name1 floor1 section1 name2 floor2 section2
--------- -------- --------- --------- --------- --------- --------- -----------
Anaheim Monday Sean HR 12 Jenny QA 24
Anaheim Monday Jenny QA 24 Sean HR 12
Cerritos Saturday Dennis Main 31 Sean HR 12
在这种情况下，我不能使用任何“排序”和删除机制..
但是，谢谢大家为解决这个问题付出的努力！
答案 0 :(得分：2)
使用聪明的GROUP BY：
SELECT t1.place, t1.day, t1.name1, t1.floor, t1.name2, t1.floor2
FROM table t1
INNER JOIN
(
SELECT t1.place, t1.day, MAX(t1.floor, t1.floor2) AS floor1,
MIN(t1.floor, t1.floor2) AS floor2
FROM table t1, table t2
GROUP BY t1.place, t1.day, MAX(t1.floor, t1.floor2),
MIN(t1.floor, t1.floor2)
) t2
ON t1.place = t2.place AND t1.day = t2.day AND t1.floor = t2.floor1 AND
t1.floor2 = t2.floor2
答案 1 :(得分：1)
首先，如果您要将floor1与9个楼层进行比较，那么2nd将大于11th。要解决此问题，您需要从floor1中提取楼层编号，然后进行比较。
话虽如此，如果它适用于您的db / version，您可以使用此查询。
通用版本就是这样的。
SQLFiddle Demo Generic verison
SELECT
t.*
FROM table1 t
INNER JOIN (SELECT
t1.place,
t1.day,
t1.name1
FROM table1 t1
INNER JOIN table1 t2
ON t1.place = t2.place
AND t1.day = t2.day
AND CAST(REPLACE(REPLACE(REPLACE(REPLACE(t1.floor1, 'st', ''), 'nd', ''), 'rd', ''), 'th', '') AS decimal)
<=
CAST(REPLACE(REPLACE(REPLACE(REPLACE(t2.floor1, 'st', ''), 'nd', ''), 'rd', ''), 'th', '') AS decimal)
GROUP BY t1.place,
t1.day,
t1.floor1
HAVING COUNT(*) = 1) t3
ON t.place = t3.place
AND t.day = t3.day
AND t.name1 = t3.name1
答案 2 :(得分：0)
使用函数LEAST和GREATEST获取已排序的名称和楼层，然后通过连接地点，日期，名称和楼层列来对数据进行分组。
试试这个：
SELECT t1.*
FROM table1 t1
GROUP BY concat(t1.place,t1.day,least(name1,name2),greatest(name1,name2),
least(floor1,floor2),greatest(floor1,floor2));