mysql - 简单的 MySQL 自连接查询，想要为每个重复的条目获取 ID 列

Question

SQL 小提琴-> http://sqlfiddle.com/#!2/28938/9

我有一张这样的桌子：

+------+----------+--------+
|  id  | group_id | letter |
+------+----------+--------+
|  1   | 44       | a      |
|  2   | 55       | a      |
|  3   | 44       | b      |
|  4   | 55       | c      |
|  5   | 44       | c      |
|  6   | 55       | d      |
+------+----------+--------+

而且我想选择任何具有两个或更多计数的字母以及具有该字母的两个 group_id。这是我正在处理的查询：

SELECT b.id AS idx, a.id AS idy, a.letter FROM (
  SELECT id, letter, COUNT(1) AS count
  FROM temp WHERE group_id = 55
  OR group_id = 44
  GROUP BY letter
) AS a
JOIN temp AS b ON a.letter = b.letter
WHERE a.count > 1
AND group_id = 55

这是我想要的结果：

+-----+-----+--------+
| idx | idy | letter |
+-----+-----+--------+
| 2   | 1   | a      |
| 4   | 5   | c      |
+-----+-----+--------+

但不幸的是，这就是我得到的

+-----+-----+--------+
| idx | idy | letter |
+-----+-----+--------+
| 2   | 1   | a      |
| 4   | 4   | c      |
+-----+-----+--------+

如何保证 ID 在正确的列中？

谢谢！

Answer 1

这不是您想要的布局，但它确实返回了 4 个字段。您上面的查询不起作用，因为您的内部查询不会导致 id #5。下面的修改后的查询可以做到，但感觉很hacky。

SELECT
      MAX(IF (b.group_id = 55, b.id, -1)) as idx,
      MAX(IF (b.group_id = 44, b.id, -1)) as idy,
      b.letter
FROM 
  temp b

INNER JOIN (
  SELECT 
      letter, group_id, COUNT(1) AS count
    FROM temp WHERE group_id = 55
    OR group_id = 44
    GROUP BY letter
    HAVING count > 1
  ) a ON (b.letter = a.letter)

GROUP BY letter