Question

我有一张这样的表：

index|from | to
------------------
  1  | ABC | DEF
  2  | ABC | GHI
  3  | ABC | GHI
  4  | ABC | JKL
  5  | ABC | JKL
  6  | ABC | JKL
  7  | DEF | ABC
  8  | DEF | GHI
  9  | DEF | JKL
 10  | GHI | ABC
 11  | GHI | ABC
 12  | GHI | ABC
 13  | JKL | DEF

我需要计算点之间的总时间（不论方向）如何得到结果：

 A  |  B  | count
-----------------
ABC | DEF | 2
ABC | GHI | 5
ABC | JKL | 3
DEF | GHI | 1
DEF | JKL | 2

到目前为止，我可以得到：

SELECT `a`.`from` as `A`, `a`.`to` as `B`, (`a`.`count` + `b`.`count`) as `count`
FROM
(SELECT `from`, `to`, count(*) as `count`  
FROM `table` 
GROUP BY 1,2) `a`
LEFT OUTER JOIN
(SELECT `from`,`to`, count(*) as `count`  
FROM `table` 
GROUP BY 1,2) `b`
ON `a`.`from` = `b`.`to` 
AND `a`.`to` = `b`.`from`

但我不确定如何删除A / B交换重复项。

Answer 1

SELECT 'from' as first, 'to' as second
FROM 'table'
UNION
SELECT 'to' as first, 'from' as second
FROM 'table'

这会给你所有的对，前进和后退。要消除重复项，您只需要一个在第一个和第二个上强加顺序的WHERE子句。

SELECT first, second, count(*)
FROM (
SELECT 'from' as first, 'to' as second
FROM 'table'
UNION
SELECT 'to' as first, 'from' as second
FROM 'table'
) x
WHERE first < second
GROUP BY first, second

避免使用字段和表名称的关键字会更简单;你可以不用引号。

Answer 2

select `from` as A, `to` as B, count(*) as `count`
from your_table
group by if (from<to, concat(from,to), concat(to,from));

获得A - ＆gt;的计数B和B-> A没有重复

2 个答案: