MySQL的 - 排除冲突的日期范围

Question

试想日这个名单范围

> 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 | /*/    */ 
+----+------------+------------+ 

我试图删除冲突的范围，只保留每个冲突组的第一次出现。

这是我应该在最后：

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

这里是fiddle

Answer 1

这里是我的可能的解决方案：

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，把结果通过一系列的范围和MySQL的容忍度的好处，以每个的第一个。

我敢肯定有更简单的

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

查询修改，以适应明显修正简短