将 2 table 与每行之间的日期映射
Mapping 2 table with between date for each row
我有 2 个 table 列，如下所示：
Table一个
rows, key_a, date_a
1, 'k1', '2015-11-12'
2, 'k2', '2015-11-20'
3, 'k3', '2015-12-01'
TableB
row, key_b, date_b, Code
1, 'k1', '2015-10-12', C1
2, 'k1', '2015-09-12', C2
3, 'k1', '2015-11-01', C3
4, 'k1', '2015-10-20', C4
5, 'k1', '2015-08-19', C5
6, 'k1', '2015-11-02', C6
7, 'k2', '2015-10-12', C7
8, 'k2', '2015-09-12', C8
9, 'k2', '2015-11-01', C9
10, 'k2', '2015-10-20', C10
11, 'k2', '2015-08-19', C11
12, 'k2', '2015-11-02', C12
13, 'k3', '2015-10-12', C13
14, 'k3', '2015-09-12', C14
15, 'k3', '2015-11-01', C15
16, 'k3', '2015-10-20', C16
17, 'k3', '2015-08-19', C17
18, 'k3', '2015-11-02', C18
我想在 table B 中找到行 date_b 是第一个具有相同 key_b 并且只有 select 数据在 table B 中具有 date_b 从 date_a 到 date_a - 30
意思是：
'k1' => '2015-11-12' to '2015-10-12' => rows view (1,3,4,6)
'k2' => '2015-11-20' to '2015-10-20' => rows view (9,10,12)
'k3' => '2015-12-01' to '2015-11-01' => rows view (15,18)
结果
key_a, date_a, Code
'k1', '2015-11-12', C1
'k2', '2015-11-20', C10
'k3', '2015-12-01', C15
我该怎么做？
所以这是你可能想要的：
select key_a,first(date_a) as first_date_a,group_concat(code) as all_codes,first(code) as first_code from (
select * from (select 1 as rows, 'k1' as key_a, date('2015-11-12') as date_a),
(select 2 as rows, 'k2' as key_a, date('2015-11-20') as date_a),
(select 3 as rows, 'k3' as key_a, date('2015-12-01') as date_a)
) table_a
join (select * from
(select 1 as rows, 'k1' as key_b, date('2015-10-12') as date_b, 'C1' as code),
(select 2 as rows, 'k1' as key_b, date('2015-09-12') as date_b, 'C2' as code),
(select 3 as rows, 'k1' as key_b, date('2015-11-01') as date_b, 'C3' as code),
(select 4 as rows, 'k1' as key_b, date('2015-10-20') as date_b, 'C4' as code),
(select 5 as rows, 'k1' as key_b, date('2015-08-19') as date_b, 'C5' as code),
(select 6 as rows, 'k1' as key_b, date('2015-11-02') as date_b, 'C6' as code),
(select 7 as rows, 'k2' as key_b, date('2015-10-12') as date_b, 'C7' as code),
(select 8 as rows, 'k2' as key_b, date('2015-09-12') as date_b, 'C8' as code),
(select 9 as rows, 'k2' as key_b, date('2015-11-01') as date_b, 'C9' as code),
(select 10 as rows, 'k2' as key_b, date('2015-10-20') as date_b, 'C10' as code),
(select 11 as rows, 'k2' as key_b, date('2015-08-19') as date_b, 'C11' as code),
(select 12 as rows, 'k2' as key_b, date('2015-11-02') as date_b, 'C12' as code),
(select 13 as rows, 'k3' as key_b, date('2015-10-12') as date_b, 'C13' as code),
(select 14 as rows, 'k3' as key_b, date('2015-09-12') as date_b, 'C14' as code),
(select 15 as rows, 'k3' as key_b, date('2015-11-01') as date_b, 'C15' as code),
(select 16 as rows, 'k3' as key_b, date('2015-10-20') as date_b, 'C16' as code),
(select 17 as rows, 'k3' as key_b, date('2015-08-19') as date_b, 'C17' as code),
(select 18 as rows, 'k3' as key_b, date('2015-11-02') as date_b, 'C18' as code)
) table_b
on table_a.key_a=table_b.key_b
where timestamp(date_b) between DATE_ADD(timestamp(date_a), -1, "MONTH") and timestamp(date_a)
group by 1
returns:
+-----+-------+--------------+-------------+------------+---+
| Row | key_a | first_date_a | all_codes | first_code | |
+-----+-------+--------------+-------------+------------+---+
| 1 | k1 | 2015-11-12 | C1,C3,C4,C6 | C1 | |
| 2 | k2 | 2015-11-20 | C9,C10,C12 | C9 | |
| 3 | k3 | 2015-12-01 | C15,C18 | C15 | |
+-----+-------+--------------+-------------+------------+---+
SELECT key_a, date_a, code
FROM (
SELECT *, ROW_NUMBER() OVER(PARTITION BY key_a ORDER BY date_b) AS num
FROM (
SELECT key_a, date_a, date_b, code,
FROM table_a AS a
JOIN table_b AS b
ON a.key_a = b.key_b
WHERE date_b BETWEEN DATE(DATE_ADD(TIMESTAMP(date_a), -1, "MONTH")) AND date_a
)
)
WHERE num = 1
ORDER BY key_a
结果：
key_a date_a code
k1 2015-11-12 C1
k2 2015-11-20 C10
k3 2015-12-01 C15
这绝对是一个 hacky 方法，但这应该可行
Select key_a, first(date_a), RIGHT(first(amalgam),3) from( select key_a, date_a, date_b + ' ' + code as amalgam from (
select * from (select 1 as rows, 'k1' as key_a, date('2015-11-12') as date_a),
(select 2 as rows, 'k2' as key_a, date('2015-11-20') as date_a),
(select 3 as rows, 'k3' as key_a, date('2015-12-01') as date_a)
) table_a
join (select * from
(select 1 as rows, 'k1' as key_b, date('2015-10-12') as date_b, 'C1' as code),
(select 2 as rows, 'k1' as key_b, date('2015-09-12') as date_b, 'C2' as code),
(select 3 as rows, 'k1' as key_b, date('2015-11-01') as date_b, 'C3' as code),
(select 4 as rows, 'k1' as key_b, date('2015-10-20') as date_b, 'C4' as code),
(select 5 as rows, 'k1' as key_b, date('2015-08-19') as date_b, 'C5' as code),
(select 6 as rows, 'k1' as key_b, date('2015-11-02') as date_b, 'C6' as code),
(select 7 as rows, 'k2' as key_b, date('2015-10-12') as date_b, 'C7' as code),
(select 8 as rows, 'k2' as key_b, date('2015-09-12') as date_b, 'C8' as code),
(select 9 as rows, 'k2' as key_b, date('2015-11-01') as date_b, 'C9' as code),
(select 10 as rows, 'k2' as key_b, date('2015-10-20') as date_b, 'C10' as code),
(select 11 as rows, 'k2' as key_b, date('2015-08-19') as date_b, 'C11' as code),
(select 12 as rows, 'k2' as key_b, date('2015-11-02') as date_b, 'C12' as code),
(select 13 as rows, 'k3' as key_b, date('2015-10-12') as date_b, 'C13' as code),
(select 14 as rows, 'k3' as key_b, date('2015-09-12') as date_b, 'C14' as code),
(select 15 as rows, 'k3' as key_b, date('2015-11-01') as date_b, 'C15' as code),
(select 16 as rows, 'k3' as key_b, date('2015-10-20') as date_b, 'C16' as code),
(select 17 as rows, 'k3' as key_b, date('2015-08-19') as date_b, 'C17' as code),
(select 18 as rows, 'k3' as key_b, date('2015-11-02') as date_b, 'C18' as code)
) table_b
on table_a.key_a=table_b.key_b
where timestamp(date_b) between DATE_ADD(timestamp(date_a), -1, "MONTH") and timestamp(date_a))
group by 1
如果你想要一个更简洁的解决方案，而不是在 date_b 和代码之间添加一个空格，你总是可以在不添加 space 的情况下添加它们，然后执行 RIGHT(2) 而不是 RIGHT (3)、取决于'amalgam'的长度。老实说，我更喜欢 Mikhail 的解决方案 :)
我有 2 个 table 列，如下所示：
Table一个
rows, key_a, date_a
1, 'k1', '2015-11-12'
2, 'k2', '2015-11-20'
3, 'k3', '2015-12-01'
TableB
row, key_b, date_b, Code
1, 'k1', '2015-10-12', C1
2, 'k1', '2015-09-12', C2
3, 'k1', '2015-11-01', C3
4, 'k1', '2015-10-20', C4
5, 'k1', '2015-08-19', C5
6, 'k1', '2015-11-02', C6
7, 'k2', '2015-10-12', C7
8, 'k2', '2015-09-12', C8
9, 'k2', '2015-11-01', C9
10, 'k2', '2015-10-20', C10
11, 'k2', '2015-08-19', C11
12, 'k2', '2015-11-02', C12
13, 'k3', '2015-10-12', C13
14, 'k3', '2015-09-12', C14
15, 'k3', '2015-11-01', C15
16, 'k3', '2015-10-20', C16
17, 'k3', '2015-08-19', C17
18, 'k3', '2015-11-02', C18
我想在 table B 中找到行 date_b 是第一个具有相同 key_b 并且只有 select 数据在 table B 中具有 date_b 从 date_a 到 date_a - 30
意思是：
'k1' => '2015-11-12' to '2015-10-12' => rows view (1,3,4,6)
'k2' => '2015-11-20' to '2015-10-20' => rows view (9,10,12)
'k3' => '2015-12-01' to '2015-11-01' => rows view (15,18)
结果
key_a, date_a, Code
'k1', '2015-11-12', C1
'k2', '2015-11-20', C10
'k3', '2015-12-01', C15
我该怎么做？
所以这是你可能想要的：
select key_a,first(date_a) as first_date_a,group_concat(code) as all_codes,first(code) as first_code from (
select * from (select 1 as rows, 'k1' as key_a, date('2015-11-12') as date_a),
(select 2 as rows, 'k2' as key_a, date('2015-11-20') as date_a),
(select 3 as rows, 'k3' as key_a, date('2015-12-01') as date_a)
) table_a
join (select * from
(select 1 as rows, 'k1' as key_b, date('2015-10-12') as date_b, 'C1' as code),
(select 2 as rows, 'k1' as key_b, date('2015-09-12') as date_b, 'C2' as code),
(select 3 as rows, 'k1' as key_b, date('2015-11-01') as date_b, 'C3' as code),
(select 4 as rows, 'k1' as key_b, date('2015-10-20') as date_b, 'C4' as code),
(select 5 as rows, 'k1' as key_b, date('2015-08-19') as date_b, 'C5' as code),
(select 6 as rows, 'k1' as key_b, date('2015-11-02') as date_b, 'C6' as code),
(select 7 as rows, 'k2' as key_b, date('2015-10-12') as date_b, 'C7' as code),
(select 8 as rows, 'k2' as key_b, date('2015-09-12') as date_b, 'C8' as code),
(select 9 as rows, 'k2' as key_b, date('2015-11-01') as date_b, 'C9' as code),
(select 10 as rows, 'k2' as key_b, date('2015-10-20') as date_b, 'C10' as code),
(select 11 as rows, 'k2' as key_b, date('2015-08-19') as date_b, 'C11' as code),
(select 12 as rows, 'k2' as key_b, date('2015-11-02') as date_b, 'C12' as code),
(select 13 as rows, 'k3' as key_b, date('2015-10-12') as date_b, 'C13' as code),
(select 14 as rows, 'k3' as key_b, date('2015-09-12') as date_b, 'C14' as code),
(select 15 as rows, 'k3' as key_b, date('2015-11-01') as date_b, 'C15' as code),
(select 16 as rows, 'k3' as key_b, date('2015-10-20') as date_b, 'C16' as code),
(select 17 as rows, 'k3' as key_b, date('2015-08-19') as date_b, 'C17' as code),
(select 18 as rows, 'k3' as key_b, date('2015-11-02') as date_b, 'C18' as code)
) table_b
on table_a.key_a=table_b.key_b
where timestamp(date_b) between DATE_ADD(timestamp(date_a), -1, "MONTH") and timestamp(date_a)
group by 1
returns:
+-----+-------+--------------+-------------+------------+---+
| Row | key_a | first_date_a | all_codes | first_code | |
+-----+-------+--------------+-------------+------------+---+
| 1 | k1 | 2015-11-12 | C1,C3,C4,C6 | C1 | |
| 2 | k2 | 2015-11-20 | C9,C10,C12 | C9 | |
| 3 | k3 | 2015-12-01 | C15,C18 | C15 | |
+-----+-------+--------------+-------------+------------+---+
SELECT key_a, date_a, code
FROM (
SELECT *, ROW_NUMBER() OVER(PARTITION BY key_a ORDER BY date_b) AS num
FROM (
SELECT key_a, date_a, date_b, code,
FROM table_a AS a
JOIN table_b AS b
ON a.key_a = b.key_b
WHERE date_b BETWEEN DATE(DATE_ADD(TIMESTAMP(date_a), -1, "MONTH")) AND date_a
)
)
WHERE num = 1
ORDER BY key_a
结果：
key_a date_a code
k1 2015-11-12 C1
k2 2015-11-20 C10
k3 2015-12-01 C15
这绝对是一个 hacky 方法，但这应该可行
Select key_a, first(date_a), RIGHT(first(amalgam),3) from( select key_a, date_a, date_b + ' ' + code as amalgam from (
select * from (select 1 as rows, 'k1' as key_a, date('2015-11-12') as date_a),
(select 2 as rows, 'k2' as key_a, date('2015-11-20') as date_a),
(select 3 as rows, 'k3' as key_a, date('2015-12-01') as date_a)
) table_a
join (select * from
(select 1 as rows, 'k1' as key_b, date('2015-10-12') as date_b, 'C1' as code),
(select 2 as rows, 'k1' as key_b, date('2015-09-12') as date_b, 'C2' as code),
(select 3 as rows, 'k1' as key_b, date('2015-11-01') as date_b, 'C3' as code),
(select 4 as rows, 'k1' as key_b, date('2015-10-20') as date_b, 'C4' as code),
(select 5 as rows, 'k1' as key_b, date('2015-08-19') as date_b, 'C5' as code),
(select 6 as rows, 'k1' as key_b, date('2015-11-02') as date_b, 'C6' as code),
(select 7 as rows, 'k2' as key_b, date('2015-10-12') as date_b, 'C7' as code),
(select 8 as rows, 'k2' as key_b, date('2015-09-12') as date_b, 'C8' as code),
(select 9 as rows, 'k2' as key_b, date('2015-11-01') as date_b, 'C9' as code),
(select 10 as rows, 'k2' as key_b, date('2015-10-20') as date_b, 'C10' as code),
(select 11 as rows, 'k2' as key_b, date('2015-08-19') as date_b, 'C11' as code),
(select 12 as rows, 'k2' as key_b, date('2015-11-02') as date_b, 'C12' as code),
(select 13 as rows, 'k3' as key_b, date('2015-10-12') as date_b, 'C13' as code),
(select 14 as rows, 'k3' as key_b, date('2015-09-12') as date_b, 'C14' as code),
(select 15 as rows, 'k3' as key_b, date('2015-11-01') as date_b, 'C15' as code),
(select 16 as rows, 'k3' as key_b, date('2015-10-20') as date_b, 'C16' as code),
(select 17 as rows, 'k3' as key_b, date('2015-08-19') as date_b, 'C17' as code),
(select 18 as rows, 'k3' as key_b, date('2015-11-02') as date_b, 'C18' as code)
) table_b
on table_a.key_a=table_b.key_b
where timestamp(date_b) between DATE_ADD(timestamp(date_a), -1, "MONTH") and timestamp(date_a))
group by 1
如果你想要一个更简洁的解决方案，而不是在 date_b 和代码之间添加一个空格，你总是可以在不添加 space 的情况下添加它们，然后执行 RIGHT(2) 而不是 RIGHT (3)、取决于'amalgam'的长度。老实说，我更喜欢 Mikhail 的解决方案