这是我以前的问题:How can I merge two strings of comma-separated numbers in MySQL?
我出于以下原因尝试使用定界列表:
我以为我别无选择。但我发现了这一点:SQL split values to multiple rows
所以,我从中获得了一些希望。但是将它应用于我的表是非常困难的。
我的示例表如下:
+-----+------+--------+------+-------+-----------+-------------+
| cid | sid | type | day | time | building | room_number |
+-----+------+--------+------+-------+-----------+-------------+
| 1 | 1 | daytime | mon | 6,7,8 | sky | 507 |
| 2 | 2 | daytime | thu | 2,3,4 | nuri | 906 |
| 3 | 3 | daytime | tue | 6,7,8 | nuri | 906 |
| 4 | 4 | daytime | thu | 6,7 | sky | 1003 |
| 5 | 5 | daytime | mon | 2,3,4 | sky | 507 |
| 6 | 6 | daytime | wed | 6,7,8 | belief | 1003 |
| 7 | 7 | daytime | mon | 2,3,4 | belief | 905 |
| 8 | 8 | daytime | fri | 6,7,8 | truth | 905 |
| 9 | 9 | daytime | tue | 6,7,8 | truth | 905 |
| 10 | 10 | daytime | fri | 2,3,4 | truth | 905 |
| 11 | 11 | daytime | wed | 6,7,8 | truth | 905 |
| 12 | 12 | daytime | fri | 2,3,4 | truth | 1003 |
| 13 | 13 | daytime | mon | 6,7,8 | truth | 905 |
| 14 | 14 | daytime | tue | 2,3,4 | truth | 905 |
| 15 | 15 | daytime | tue | 6,7,8 | sky | 208 |
| 16 | 16 | daytime | tue | 2,3,4 | sky | 208 |
| 17 | 17 | daytime | tue | 2,3,4 | truth | 1004 |
| 18 | 19 | daytime | mon | 2,3,4 | sky | 208 |
| 19 | 20 | daytime | thu | 2,3,4 | truth | 1003 |
| 20 | 21 | daytime | wed | 6,7,8 | sky | 208 |
| 21 | 22 | night | tue | 4,5,6 | nuri | 405 |
| 22 | 23 | night | tue | 1,2,3 | nuri | 405 |
| 23 | 24 | night | tue | 1,2,3 | nuri | 306 |
| 24 | 25 | night | thu | 1,2,3 | nuri | 205 |
| 25 | 26 | night | thu | 4,5,6 | sky | 306 |
| 26 | 27 | night | wed | 1,2,3 | nuri | 306 |
| 27 | 28 | night | wed | 4,5,6 | sky | 309 |
| 28 | 29 | night | wed | 4,5,6 | nuri | 407 |
| 29 | 30 | night | tue | 4,5,6 | nuri | 306 |
| 30 | 31 | night | thu | 1,2,3 | nuri | 307 |
| 31 | 0 | always | | | sky | 201 |
| 32 | 0 | always | | | sky | 202 |
| 33 | 0 | always | | | sky | 203 |
| 34 | 0 | always | | | sky | 204 |
| 35 | 0 | always | | | nuri | 205 |
| 36 | 0 | always | | | nuri | 206 |
| 37 | 0 | always | | | truth | 207 |
| ... | ... | ... | | | ... | ... |
| 2000 | 0 | always | | | belief | 1101 |
+-----+------+--------+------+-------+-----------+-------------+
我对更改表的期望是这样:
+-----+------+--------+------+-------+-----------+-------------+
| cid | sid | type | day | time | building | room_number |
+-----+------+--------+------+-------+-----------+-------------+
| 1 | 1 | daytime | mon | 1 | sky | 507 |
| 2 | 1 | daytime | mon | 5 | sky | 507 |
| 3 | 1 | daytime | mon | 9 | sky | 507 |
| 4 | 1 | daytime | mon | 10 | sky | 507 |
| 5 | 1 | daytime | mon | 11 | sky | 507 |
| 6 | 1 | daytime | mon | 12 | sky | 507 |
| 7 | 1 | daytime | mon | 13 | sky | 507 |
| 8 | 2 | daytime | thu | 1 | nuri | 906 |
| 9 | 2 | daytime | thu | 5 | nuri | 906 |
| 10 | 2 | daytime | thu | 6 | nuri | 906 |
| 11 | 2 | daytime | thu | 7 | nuri | 906 |
| 12 | 2 | daytime | thu | 8 | nuri | 906 |
| 13 | 2 | daytime | thu | 9 | nuri | 906 |
| 14 | 2 | daytime | thu | 10 | nuri | 906 |
| 15 | 2 | daytime | thu | 11 | nuri | 906 |
| 16 | 2 | daytime | thu | 12 | nuri | 906 |
| 17 | 2 | daytime | thu | 13 | nuri | 906 |
| 18 | 3 | daytime | tue | 1 | nuri | 906 |
| 19 | 3 | daytime | tue | 2 | nuri | 906 |
| 20 | 3 | daytime | tue | 3 | nuri | 906 |
| ... | ... | ... | | | ... | ... |
| 302 | 0 | always | | | nuri | 206 |
| 303 | 0 | always | | | truth | 207 |
| ... | ... | ... | | | ... | ... |
| 4020 | 0 | always | | | belief | 1101 |
+-----+------+--------+------+-------+-----------+-------------+
我这样做的原因是源数据是课程的时间表。而且我想找到空余时间,以便人们在上课的时候可以使用教室。
这是一种使用回溯查询将字符串拆分为行的方法。然后,您可以使用此结果集来识别丢失的记录:为此,您可以使用时间列表来cross join
原始表,并使用反left join
来提取丢失的记录:
with recursive
cte as (
select cid, sid, type, day, room_number,
substring(time, 1, locate(',', time) - 1) time,
substring(concat(time, ','), locate(',', time) + 1) rest
from mytable
union all
select cid, sid, type, day, room_number,
substring(rest, 1, locate(',', rest) - 1),
substring(rest, locate(',', rest) + 1)
from cte
where locate(',', rest) > 0
),
all_times as (
select 1 time
union all select time + 1 from all_times where time < 13
)
select t.cid, t.sid, t.type, t.day, at.time, t.building, t.room_number
from all_times at
cross join mytable t
left join cte c on c.cid = t.cid and c.time = at.time
where c.cid is null
order by t.cid, at.time
样本数据:
cid | sid |类型天时间|建筑|房间号-:| -:| :------ | :-| :---- | :------- | ----------: 1 | 1 |白天|星期一| 6,7,8 |天空| 507 2 | 2 |白天|周四| 2,3,4 |努里906
查询结果:
cid | sid |类型天时间|建筑|房间号-:| -:| :------ | :-| ---: :------- | ----------: 1 | 1 |白天|星期一| 1 |天空| 507 1 | 1 |白天|星期一| 2 |天空| 507 1 | 1 |白天|星期一| 3 |天空| 507 1 | 1 |白天|星期一| 4 |天空| 507 1 | 1 |白天|星期一| 5 |天空| 507 1 | 1 |白天|星期一| 9 |天空| 507 1 | 1 |白天|星期一| 10 |天空| 507 1 | 1 |白天|星期一| 11 |天空| 507 1 | 1 |白天|星期一| 12 |天空| 507 1 | 1 |白天|星期一| 13 |天空| 507 2 | 2 |白天|周四| 1 |努里906 2 | 2 |白天|周四| 5 |努里906 2 | 2 |白天|周四| 6 |努里906 2 | 2 |白天|周四| 7 |努里906 2 | 2 |白天|周四| 8 |努里906 2 | 2 |白天|周四| 9 |努里906 2 | 2 |白天|周四| 10 |努里906 2 | 2 |白天|周四| 11 |努里906 2 | 2 |白天|周四| 12 |努里906 2 | 2 |白天|周四| 13 |努里906
通过在前面加上insert
指令,您可以轻松地将其转换为insert
查询,例如:
insert into sometable(cid, sid, type, day, time, building, room_number)
with recursive ...
select ...
from ...
where ...
实际上,由于csv列表中的值列表是预定义的,因此可以很好地完成而无需递归。您可以将天数列表与表格合并,然后使用find_in_set()
识别差距。好处是它可以在不支持追溯查询的MySQL / MariaDB版本上运行:
select t.cid, t.sid, t.type, t.day, at.time, t.building, t.room_number
from (
select 1 time union all select 2 time union all select 3
union all select 4 union all select 5 union all select 6
union all select 7 union all select 8 union all select 9
union all select 10 union all select 11 union all select 12 union all select 13
) at
cross join (
select distinct cid, sid, type, day, building, room_number
from mytable
) t
left join mytable t1 on t1.cid = t.cid and find_in_set(at.time, t1.time)
where t1.cid is null
order by t.cid, at.time