我有一个带有日期字段的表。我需要查询group_concat日期,只有它们是连续的,否则单独返回它们。
因此,例如,如果我的表具有以下数据:
+---------+--------------+
| user_id | checkin_date |
+---------+--------------+
| 1 | 2012-02-01 |
| 2 | 2012-03-01 |
| 3 | 2012-02-03 |
| 4 | 2012-02-02 |
+---------+--------------+
我需要一个返回以下结果的查询
+--------------------------+
| checkin_period |
+--------------------------+
| 2012-02-01 - 2012-02-03 |
| 2012-03-01 |
+--------------------------+
正如你所看到的,第1代,第2代和第3代被分为1行(仅显示第一天和最后一天),而3月1日本身就是......
我不知道从哪里开始!
提前致谢,
阿兰
答案 0 :(得分:1)
只有当checkin_date与前一行不连续时,即只要日期增加超过1天,您才可以增加用户变量@p
。
SELECT IF(checkin_date <= @d + INTERVAL 1 DAY, @p, @p:=@p+1) AS p, @d:=checkin_date AS d
FROM (SELECT @p:=0, @d:='1900-01-01') _init, mytable
ORDER BY checkin_date;
+------+------------+
| p | d |
+------+------------+
| 1 | 2012-02-01 |
| 1 | 2012-02-02 |
| 1 | 2012-02-03 |
| 2 | 2012-03-01 |
+------+------------+
然后使用上面的子查询,按p
列进行分组,并返回min到max范围的字符串,如果组内的计数为1,则只返回一个值。
SELECT IF(COUNT(*) > 1, CONCAT(MIN(d), ' - ', MAX(d)), MAX(d)) AS date_range
FROM (
SELECT IF(checkin_date <= @d + INTERVAL 1 DAY, @p, @p:=@p+1) AS p, @d:=checkin_date AS d
FROM (SELECT @p:=0, @d:='1900-01-01') _init, mytable
ORDER BY checkin_date) AS t
GROUP BY p;
+-------------------------+
| date_range |
+-------------------------+
| 2012-02-01 - 2012-02-03 |
| 2012-03-01 |
+-------------------------+
答案 1 :(得分:1)
SELECT
CONCAT_WS(' - ',
MIN(checkin_date),
CASE WHEN MAX(checkin_date)>MIN(checkin_date) THEN MAX(checkin_date) END
) As time_interval
FROM (
SELECT
CASE WHEN checkin_date=@last_ci+INTERVAL 1 DAY THEN @n ELSE @n:=@n+1 END AS g,
@last_ci := checkin_date As checkin_date
FROM
tablename, (SELECT @n:=0) r
ORDER BY
checkin_date
) s
GROUP BY
g
请参阅小提琴here。