SQL / Teradata：返回记录，其中连续行中的值相同

Question

I have a data set that looks like:

ID        date     emp_num    loc
1111     5/2/16    111111     Brooklyn
1112     5/3/16    222222     Detroit
1113     5/3/16    333333     San Diego
1114     5/2/16    333333     Orlando
1115     5/5/16    333333     Brooklyn
1116     5/7/16    111111     Orlando

In this case, I would want to return records 1113, 1114, and 1115 because the emp_num in consecutive rows (ordered by ID) is the same.

I use Teradata, but if anyone has a SQL solution for another engine I can usually manage to translate it.

Thank you.

Answer 1

You need to look at the previous/next row and check if it didn't change:

SELECT * 
FROM tab
QUALIFY 
   MIN(emp_num) --previous row
   OVER (ORDER BY ID
         ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) = emp_num
OR
   MIN(emp_num) -- next row
   OVER (ORDER BY ID
         ROWS BETWEEN 1 FOLLOWING AND 1 FOLLOWING) = emp_num

In Standard SQL this would be a task for LAG / LEAD , but Teradata doesn't impement it, so you have to rewrite it.

Answer 2

First, get the rownumber difference ordered by id column and partitioned by emp_num and ordered by id column. This would classify emp_num into groups. Then, get the groups which have more than one member in them (which means there are consecutive rows with the same emp_num value). Finally select the required columns for those groups.

WITH x AS (SELECT
  *,
  ROW_NUMBER() OVER (ORDER BY id) - ROW_NUMBER() OVER (PARTITION BY emp_num ORDER BY id) grp
FROM t),
grpsneeded
AS (SELECT
  grp
FROM x
GROUP BY grp
HAVING COUNT(*) > 1)
SELECT
  id,
  dt,
  emp_num
FROM x
WHERE grp IN (SELECT
  grp
FROM grpsneeded)

Sample Demo

This solution works well with SQL Server.

A more simpler SQL solution would be using lead and lag functions. As @dnoeth pointed out, Teradata doesn't support these functions. However, this may be useful for other database engines.

select id, dt , emp_num from (
select *
,lead(emp_num) over(order by id) nxt
,lag(emp_num) over(order by id) prev
from t
) x
where coalesce(nxt,0) = emp_num or coalesce(prev,0) = emp_num