警告：没有写入/运行PROC的权限，也无法创建/修改临时表。
使用字段'unique_primary_key'，'cust_num'和'sale_date'给多字节表'table1'。 对于'cust_num'的每个不同值，存在多个记录，其值为'sale_date'。
我需要：在通过'cust_num'进行分区后，找到有问题记录的'sale_date'与分区内每隔一条记录的'sale_date'之间差异的最小值。此外，该差异的第二个最小值，第三个最小值。
每个不同的'cust_num'在表格中有3到75个记录（所有记录都有不同的日期），所以只需在分区内按'sale_date'排序，然后继续明确地将'sale_date'值与其他所有记录进行比较分区不可行。
我知道如何使用INDEX函数和MATCH作为参数之一等在Excel中轻松完成此操作，但我不知道SQL中的类似过程。
答案 0 :(得分：0)
如果你只想要一个前瞻性或后视的日期差异衡量标准，我认为你可以采用简单的滞后/牵头方法来做到这一点。
select * from table1 order by cust_num, sale_date;
UNIQUE_PRIMARY_KEY | CUST_NUM | SALE_DATE
--------------------+----------+------------
7 | 1 | 2014-10-06
2 | 1 | 2014-10-12
5 | 1 | 2014-10-14
1 | 1 | 2014-10-17
4 | 1 | 2014-10-19
3 | 1 | 2014-10-22
6 | 1 | 2014-10-25
100008 | 5 | 2014-10-07
300002 | 5 | 2014-10-13
100006 | 5 | 2014-10-15
200003 | 5 | 2014-10-18
200004 | 5 | 2014-10-20
100005 | 5 | 2014-10-23
100007 | 5 | 2014-10-26
则...
SELECT UNIQUE_PRIMARY_KEY pk,
CUST_NUM,
SALE_DATE,
lead(sale_date,1) over (partition BY cust_num ORDER BY sale_date) lead1,
lead(sale_date,2) over (partition BY cust_num ORDER BY sale_date) lead2,
lead(sale_date,3) over (partition BY cust_num ORDER BY sale_date) lead3
FROM table1
ORDER BY CUST_NUM,
SALE_DATE
PK | CUST_NUM | SALE_DATE | LEAD1 | LEAD2 | LEAD3
--------+----------+------------+------------+------------+------------
7 | 1 | 2014-10-06 | 2014-10-12 | 2014-10-14 | 2014-10-17
2 | 1 | 2014-10-12 | 2014-10-14 | 2014-10-17 | 2014-10-19
5 | 1 | 2014-10-14 | 2014-10-17 | 2014-10-19 | 2014-10-22
1 | 1 | 2014-10-17 | 2014-10-19 | 2014-10-22 | 2014-10-25
4 | 1 | 2014-10-19 | 2014-10-22 | 2014-10-25 |
3 | 1 | 2014-10-22 | 2014-10-25 | |
6 | 1 | 2014-10-25 | | |
100008 | 5 | 2014-10-07 | 2014-10-13 | 2014-10-15 | 2014-10-18
300002 | 5 | 2014-10-13 | 2014-10-15 | 2014-10-18 | 2014-10-20
100006 | 5 | 2014-10-15 | 2014-10-18 | 2014-10-20 | 2014-10-23
200003 | 5 | 2014-10-18 | 2014-10-20 | 2014-10-23 | 2014-10-26
200004 | 5 | 2014-10-20 | 2014-10-23 | 2014-10-26 |
100005 | 5 | 2014-10-23 | 2014-10-26 | |
100007 | 5 | 2014-10-26 | | |
(14 rows)
在更可能的情况下，你希望它是日期差异的ABS，我的大脑仍然想要使用滞后＆amp;领导，但是它正在努力摆脱它的悲伤能力，并且必须以更加复杂的方式将它拼凑在一起：
WITH foo AS
(
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lead(sale_date,1) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lead(sale_date,1) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
UNION ALL
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lead(sale_date,2) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lead(sale_date,2) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
UNION ALL
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lead(sale_date,3) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lead(sale_date,3) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
UNION ALL
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lag(sale_date,1) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lag(sale_date,1) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
UNION ALL
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lag(sale_date,2) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lag(sale_date,2) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
UNION ALL
SELECT UNIQUE_PRIMARY_KEY pk,
cust_num,
sale_date,
lag(sale_date,3) over (partition BY cust_num ORDER BY sale_date) rel_sales_date,
ABS(sale_date - lag(sale_date,3) over (partition BY cust_num ORDER BY sale_date)) day_delta
FROM table1
)
SELECT pk,
cust_num,
sale_date,
rel_sales_date,
day_delta,
day_rank
FROM (
SELECT pk ,
cust_num,
sale_date,
rel_sales_date,
day_delta,
dense_rank() over (partition BY pk ORDER BY day_delta nulls last) day_rank
FROM foo
)
foob
WHERE day_rank <= 3
ORDER BY cust_num,
sale_date,
day_rank
PK | CUST_NUM | SALE_DATE | REL_SALES_DATE | DAY_DELTA | DAY_RANK
--------+----------+------------+----------------+-----------+----------
7 | 1 | 2014-10-06 | 2014-10-12 | 6 | 1
7 | 1 | 2014-10-06 | 2014-10-14 | 8 | 2
7 | 1 | 2014-10-06 | 2014-10-17 | 11 | 3
2 | 1 | 2014-10-12 | 2014-10-14 | 2 | 1
2 | 1 | 2014-10-12 | 2014-10-17 | 5 | 2
2 | 1 | 2014-10-12 | 2014-10-06 | 6 | 3
5 | 1 | 2014-10-14 | 2014-10-12 | 2 | 1
5 | 1 | 2014-10-14 | 2014-10-17 | 3 | 2
5 | 1 | 2014-10-14 | 2014-10-19 | 5 | 3
1 | 1 | 2014-10-17 | 2014-10-19 | 2 | 1
1 | 1 | 2014-10-17 | 2014-10-14 | 3 | 2
1 | 1 | 2014-10-17 | 2014-10-12 | 5 | 3
1 | 1 | 2014-10-17 | 2014-10-22 | 5 | 3
4 | 1 | 2014-10-19 | 2014-10-17 | 2 | 1
4 | 1 | 2014-10-19 | 2014-10-22 | 3 | 2
4 | 1 | 2014-10-19 | 2014-10-14 | 5 | 3
3 | 1 | 2014-10-22 | 2014-10-25 | 3 | 1
3 | 1 | 2014-10-22 | 2014-10-19 | 3 | 1
3 | 1 | 2014-10-22 | 2014-10-17 | 5 | 2
3 | 1 | 2014-10-22 | 2014-10-14 | 8 | 3
6 | 1 | 2014-10-25 | 2014-10-22 | 3 | 1
6 | 1 | 2014-10-25 | 2014-10-19 | 6 | 2
6 | 1 | 2014-10-25 | 2014-10-17 | 8 | 3
100008 | 5 | 2014-10-07 | 2014-10-13 | 6 | 1
100008 | 5 | 2014-10-07 | 2014-10-15 | 8 | 2
100008 | 5 | 2014-10-07 | 2014-10-18 | 11 | 3
300002 | 5 | 2014-10-13 | 2014-10-15 | 2 | 1
300002 | 5 | 2014-10-13 | 2014-10-18 | 5 | 2
300002 | 5 | 2014-10-13 | 2014-10-07 | 6 | 3
100006 | 5 | 2014-10-15 | 2014-10-13 | 2 | 1
100006 | 5 | 2014-10-15 | 2014-10-18 | 3 | 2
100006 | 5 | 2014-10-15 | 2014-10-20 | 5 | 3
200003 | 5 | 2014-10-18 | 2014-10-20 | 2 | 1
200003 | 5 | 2014-10-18 | 2014-10-15 | 3 | 2
200003 | 5 | 2014-10-18 | 2014-10-13 | 5 | 3
200003 | 5 | 2014-10-18 | 2014-10-23 | 5 | 3
200004 | 5 | 2014-10-20 | 2014-10-18 | 2 | 1
200004 | 5 | 2014-10-20 | 2014-10-23 | 3 | 2
200004 | 5 | 2014-10-20 | 2014-10-15 | 5 | 3
100005 | 5 | 2014-10-23 | 2014-10-20 | 3 | 1
100005 | 5 | 2014-10-23 | 2014-10-26 | 3 | 1
100005 | 5 | 2014-10-23 | 2014-10-18 | 5 | 2
100005 | 5 | 2014-10-23 | 2014-10-15 | 8 | 3
100007 | 5 | 2014-10-26 | 2014-10-23 | 3 | 1
100007 | 5 | 2014-10-26 | 2014-10-20 | 6 | 2
100007 | 5 | 2014-10-26 | 2014-10-18 | 8 | 3
(46 rows)
希望这有帮助，如果不是，我至少在本周五下午进行一些心理锻炼。