Mysql实战之求出缺失范围
1.需求
求出缺失范围
2.示例
根据表中某个变化字段，求出变化字段的缺失范围。如下给出一个例子：
下面是一个表x，其中的数据如下：
mysql> select * from x;
+------+
| a |
+------+
| 3 |
| 4 |
| 5 |
| 6 |
| 100 |
| 101 |
| 102 |
| 105 |
| 106 |
| 107 |
+------+
10 rows in set (0.03 sec)其中的缺失范围就是：7-99,103-104。
3.代码一
- sql 1
下面这个SQL是用于查找不连续的值。【如果x1.a+1=x2.a，那么说明x1.a是连续的，反之不连续。因为这里使用的是not exists，所以就是不连续的，可以查看得到最后的结果集就是6,102,107】
select a
from x as x1
where not exists
(
select *
from x as x2
where x1.a + 1 = x2.a
);执行结果如下：
+------+
| a |
+------+
| 6 |
| 102 |
| 107 |
+------+
3 rows in set (0.00 sec)- sql 2
因为我们需要求“缺失序列”，所以由上面的结果集可知，6,102,107均非缺失值，但是将其分别+1就得到了缺失值。但是又因为107+1已经超过了表中的最大值，所以丢弃。那么可得到下述的SQL：
select a+1 as start_range
from x as x1
where not exists
(
select *
from x as x2
where x1.a + 1 = x2.a
)
and a< (select max(a) from x);执行结果如下：
+-------------+
| start_range |
+-------------+
| 7 |
| 103 |
+-------------+
2 rows in set (0.00 sec)- sql 3
因为要得到一个缺失序列，肯定要有缺失字段的结束范围，所以再找出结束值。【这里的结束值就是大于连续值的最小值-1】
select
a+1 as start_range
,(
select
min(a) -1
from x as x3
where x3.a > x1.a
) as end_range
from x as x1
where not exists
(
select a
from x as x2
where x1.a + 1 = x2.a
)
and a< (select max(a) from x);执行结果如下：
+-------------+-----------+
| start_range | end_range |
+-------------+-----------+
| 7 | 99 |
| 103 | 104 |
+-------------+-----------+
2 rows in set (0.00 sec)代码二
上述只是一个普通的解决方案，如果追求效率，可以参考下面这个更好的SQL：
select a as cur,
(
select
min(a)
from x as x1
where x1.a > x2.a
) as next
from x as x2;执行结果如下：
+------+------+
| cur | next |
+------+------+
| 3 | 4 |
| 4 | 5 |
| 5 | 6 |
| 6 | 100 |
| 100 | 101 |
| 101 | 102 |
| 102 | 105 |
| 105 | 106 |
| 106 | 107 |
| 107 | NULL |
+------+------+
10 rows in set (0.00 sec)select
cur+1 as start_range,
next-1 as end_range
from
(
select
a as cur,
(select min(a) from x as x2
where x2.a > x1.a)as next
from x as x1
)as x3
where next - cur > 1
+-------------+-----------+
| start_range | end_range |
+-------------+-----------+
| 7 | 99 |
| 103 | 104 |
+-------------+-----------+
2 rows in set (0.01 sec)
