'Oracle SQL LAG() function results in duplicate rows
I have a very simple query that results in two rows:
SELECT DISTINCT
id,
trunc(start_date) start_date
FROM example.table
WHERE ID = 1
This results in the following rows:
id start_date
1 7/1/2012
1 9/1/2016
I want to add a column that simply shows the previous date for each row. So I'm using the following:
SELECT DISTINCT id,
Trunc(start_date) start_date,
Lag(start_date, 1)
over (
ORDER BY start_date) pdate
FROM example.table
WHERE id = 1
However, when I do this, I get four rows instead of two:
id start_date pdate
1 7/1/2012 NULL
1 7/1/2012 7/1/2012
1 9/1/2016 7/1/2012
1 9/1/2016 9/1/2012
If I change the offset to 2 or 3 the results remain the same. If I change the offset to 0, I get two rows again but of course now the start_date == pdate.
I can't figure out what's going on
Solution 1:[1]
Use an explicit GROUP BY
instead:
SELECT id, trunc(start_date) as start_date,
LAG(trunc(start_date)) OVER (PARTITION BY id ORDER BY trunc(start_date))
FROM example.table
WHERE ID = 1
GROUP BY id, trunc(start_date)
Solution 2:[2]
The reason for this is: the order of execution of an SQL statements, is that LAG
runs before the DISTINCT
.
You actually want to run the LAG
after the DISTINCT
, so the right query should be:
WITH t1 AS (
SELECT DISTINCT id, trunc(start_date) start_date
FROM example.table
WHERE ID = 1
)
SELECT *, LAG(start_date, 1) OVER (ORDER BY start_date) pdate
FROM t1
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Gordon Linoff |
Solution 2 |