Question

我希望从我的表中获得两个连续值之间的差异。

| id | count |
| 1  |   1   |
| 2  |   7   |
| 3  |   9   |
| 4  |   3   |
| 5  |   7   |
| 6  |   9   |

例如

之间的区别

id2-id1 = 6，
id3-id2 = -2 ，
...

我该怎么办？ SELECT SUM(id(x+1) - id(x)) FROM table1

Answer 1

如果您知道 ids没有间隙，那么只需使用join：

select t.*, (tnext.count - t.count) as diff
from table t join
     table tnext
     on t.id = tnext.id - 1;

如果您只想要差异的总和，那么这与最后一个值减去第一个值相同（所有中间值在求和中抵消）。您可以使用limit：

执行此操作

select last.count - first.count
from (select t.* from table order by id limit 1) as first cross join
     (select t.* from table order by id desc limit 1) as last;

Answer 2

您可以使用子查询查找前一个count的{{1}}。

如果id列中没有空白：

ID

SQLFiddle

如果SELECT CONCAT(t.`id` ,' - ', t.`id` - 1) AS `IDs` , t.`count` - (SELECT `count` FROM `tbl` WHERE `id` = t.`id` - 1) AS `Difference` FROM `tbl` t WHERE t.`id` > 1列中存在空白。 第一个解决方案，ID使用ORDER BY <...> DESC：

LIMIT 1

SQLFiddle

第二个解决方案，使用其他子查询查找SELECT CONCAT(t.id ,' - ', (SELECT `id` FROM tbl WHERE t.id > id ORDER BY id DESC LIMIT 1)) AS IDs , t.`count` - (SELECT `count` FROM tbl WHERE t.id > id ORDER BY id DESC LIMIT 1) AS difference FROM tbl t WHERE t.id > 1;，count小于当前MAX(id)：

id

SQLFiddle

P.S。：第一列SELECT CONCAT(t.id ,' - ', (SELECT MAX(`id`) FROM tbl WHERE id < t.id)) AS IDs , t.`count` - (SELECT `count` FROM tbl WHERE `id` = (SELECT MAX(`id`) FROM tbl WHERE id < t.id) ) AS difference FROM tbl t WHERE t.id > 1;仅用于提高可读性，如果有必要，您可以省略或完全更改。

Answer 3

试试这个：

SELECT MAX(count)-MIN(count) diff WHERE id IN(1,2)

或者这样

SELECT 2*STD(count) diff WHERE id IN(1,2)

Answer 4

即使 ID之间有距离：

，这也有效

SELECT *, 
  ((SELECT value FROM example e2 WHERE e2.id > e1.id ORDER BY id ASC LIMIT 1) - value) as diff
FROM example e1;

Mysql - 获取两个连续值之间的差异

4 个答案: