Question

我有以下示例记录：

double

当存在重复的employee_id时，如何过滤log_id = 12345、12346（log_id仅相隔1个数字）的记录？输出应为：

log_id employee_id
12345  99999      
12346  99999      
12347  88888      
12357  88888

Answer 1

我不会使用窗口功能。我只会使用import itertools import datetime from dateutil.parser import parse transactions = ['2018-12-04 13:{}0:00+00:00'.format(i) for i in range(6)] + \ ['2018-12-04 14:{}0:00+00:00'.format(i) for i in range(1)] + \ ['2018-12-04 15:{}0:00+00:00'.format(i) for i in range(2)] for timestamp, grp in itertools.groupby(transactions, key=lambda x: datetime.datetime.combine(parse(x).date(), datetime.time(parse(x).hour, 0, 0, 0))): count = list(grp) print('{}:{}'.format(timestamp, len(count)))：

2018-12-04 13:00:00:6
2018-12-04 14:00:00:1
2018-12-04 15:00:00:2

该查询应该能够利用exists上的索引。

Answer 2

我会尝试这样的事情：

with 
x as (
  select log_id, employee_id, row_number() over(order by log_id) as rn
  from my_table
),
y as (
  select
    log_id, employee_id, rn,
    lag(log_id) over(order by rn) as prev_log_id,
    lead(log_id) over(order by rn) as next_log_id
from x
)
select log_id, employee_id from y
where log_id - 1 = prev_log_id or prev_log_id is null
  and log_id + 1 = next_log_id or next_log_is is null
order by rn

Answer 3

另一个选择：

SQL> with test (log_id, employee_id) as
  2    (select 12345, 99999 from dual union all
  3     select 12346, 99999 from dual union all
  4     select 12344, 99999 from dual union all  --> added this one
  5     --
  6     select 12347, 88888 from dual union all
  7     select 12357, 88888 from dual
  8    )
  9  select log_id, employee_id
 10  from test
 11  where employee_id in (select employee_id
 12                        from test
 13                        group by employee_id
 14                        having max(log_id) - min(log_id) = count(*) - 1
 15                       );

    LOG_ID EMPLOYEE_ID
---------- -----------
     12344       99999
     12346       99999
     12345       99999

SQL>

Answer 4

假设此表结构：

create table test (
  log_id number(5) primary key,
  employee_id number(5) not null
)

此示例数据：

insert into test
   (select 12342, 99999 from dual union all
    select 12343, 77777 from dual union all
    select 12344, 99999 from dual union all
    select 12345, 99999 from dual union all
    select 12346, 99999 from dual union all
    select 12347, 88888 from dual union all
    -- gap
    select 12357, 88888 from dual union all
    select 12358, 33333 from dual union all
    select 12359, 33333 from dual
   )

您可以通过以下查询来做到这一点：

with x as (
 select log_id,
        employee_id,
        lead(log_id) over (order by log_id) as next_log_id,
        lag(log_id) over (order by log_id) as previous_log_id,
        lead(employee_id) over (order by log_id) as next_employee_id,
        lag(employee_id) over (order by log_id) as previous_employee_id
 from test
)
select log_id, employee_id
  from x
 where (log_id = next_log_id - 1 and employee_id = next_employee_id)
    or (log_id = previous_log_id + 1 and employee_id = previous_employee_id)
 order by 1

结果如下：

LOG_ID | EMPLOYEE_ID
-------+------------
 12344 |       99999
 12345 |       99999
 12346 |       99999
 12358 |       33333
 12359 |       33333

如果可以保证LOG_ID的值序列没有间隙（因为样本的范围是12342到12347），则可以使用更简单的变体：

with x as (
 select log_id,
        employee_id,
        lead(employee_id) over (order by log_id) as next_employee_id,
        lag(employee_id) over (order by log_id) as previous_employee_id
 from test
 where log_id between 12342 and 12347
)
select log_id, employee_id
  from x
 where employee_id in (previous_employee_id, next_employee_id)
 order by 1

您可以在this Oracle LiveSQL或此SQL Fiddle上看到它的运行情况。

Oracle 12c-在窗口中选择记录

4 个答案: