Oracle 12c-在窗口中选择记录

Question

I have the following sample records:

log_id employee_id
12345  99999      
12346  99999      
12347  88888      
12357  88888 

How do I filter records where log_id = 12345, 12346 (log_id only 1 number apart) when there duplicate employee_id's? Output should be:

log_id employee_id
12345  99999      
12346  99999 

Answer 1

I would not use window functions. I would simply use exists :

select t.*
from t
where exists (select 1
              from t tnext
              where tnext.log_id = t.log_id + 1 and tnext.employee_id = t.employee_id) or
      exists (select 1
              from t tprev
              where tprev.log_id = t.log_id - 1 and tprev.employee_id = t.employee_id);

This query should be able to take advantage of an index on (employee_id, log_id) .

Answer 2

I would try something like this:

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

Yet another option:

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

Assuming this table structure:

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

This sample data:

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
   )

You could do it with this query:

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

With this result:

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

If you are guaranteed, to have the LOG_ID value sequence without gaps (as the sample has in range from 12342 to 12347), you could use a simpler variant:

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

You can see it in action at this Oracle LiveSQL or this SQL Fiddle .