MySQL的 - 排除衝突的日期範圍

Question

試想日這個名單範圍

> SELECT * FROM range_table; 
+----+------------+------------+ 
| id | start  | end  | 
+----+------------+------------+ 
| 1 | 2014-01-01 | 2014-01-15 | /* -- Not conflicting */ 
| 2 | 2014-01-15 | 2014-01-16 | /* -- Conflicting  */ 
| 3 | 2014-01-15 | 2014-01-20 | /* |     */ 
| 4 | 2014-01-15 | 2014-01-19 | /*/    */ 
| 5 | 2014-01-24 | 2014-01-26 | /* -- Conflicting  */ 
| 6 | 2014-01-21 | 2014-01-25 | /*/    */ 
+----+------------+------------+ 

我試圖刪除衝突的範圍，只保留每個衝突組的第一次出現。

這是我應該在最後：

+----+------------+------------+ 
| id | start  | end  | 
+----+------------+------------+ 
| 1 | 2014-01-01 | 2014-01-15 | 
| 2 | 2014-01-15 | 2014-01-16 | 
| 5 | 2014-01-24 | 2014-01-26 | 
+----+------------+------------+ 

這裏是fiddle

Answer 1

這裏是我的可能的解決方案：

SELECT `id`, `start`, `end` 
FROM (
    SELECT 
    r1.id 
    , r1.start 
    , r1.end 
    , COUNT(DISTINCT r2.id) AS "conflicts" 
    , MD5(GROUP_CONCAT(DISTINCT r2.id ORDER BY r2.id)) AS "group_chksum" 
    FROM range_table AS r1 
    LEFT JOIN range_table AS r2 
    ON (r1.end > r2.start AND r1.start < r2.end) 
    GROUP BY r1.id 
) AS tmp 
GROUP BY group_chksum 
; 

的想法是，把結果通過一系列的範圍和MySQL的容忍度的好處，以每個的第一個。

我敢肯定有更簡單的

Answer 2

SELECT x.* 
    FROM range_table x 
    LEFT 
    JOIN range_table y 
    ON y.start < x.end 
    AND y.end > x.start 
    AND y.id < x.id 
WHERE y.id IS NULL; 

http://sqlfiddle.com/#!2/6f723/26

查詢修改，以適應明顯修正簡短