SQL: How to return just 1 previous date for a record, not all previous dates - sql

Question

I have a very simple table of ID's and Sign-in dates and I want to use SQL to make a column that shows the previous sign-in date:
Table: SIGNIN
| ID | Sign-in Date |
| A | 01/01/19 |
| B | 01/01/19 |
| C | 02/01/19 |
| A | 02/01/19 |
| A | 03/01/19 |
| B | 03/01/19 |
| A | 04/01/19 |
| C | 04/01/19 |
| B | 05/01/19 |
I've tried doing a join to itself but it's showing all previous sign-in dates rather than just the most recent.
SELECT [SIGNIN].ID
[SIGNIN].SignInDate
FROM [SIGNIN]
INNER JOIN [SIGNIN] as [Prev] on [SIGNIN].ID = [Prev].ID
and [SIGNIN].SignInDate < [Prev].SignInDate
ORDER BY [SIGNIN].ID, [SIGNIN].SignInDate
The result I want:
Table: SIGNIN
| ID | Sign-in Date | Previous |
| A | 01/01/19 | NULL |
| B | 01/01/19 | NULL |
| C | 02/01/19 | NULL |
| A | 02/01/19 | 01/01/19 |
| A | 03/01/19 | 02/01/19 |
| B | 03/01/19 | 01/01/19 |
| A | 04/01/19 | 03/01/19 |
| C | 04/01/19 | 02/01/19 |
| B | 05/01/19 | 03/01/19 |
What I'm getting:
| ID | Sign-in Date | Previous |
| A | 01/01/19 | NULL |
| B | 01/01/19 | NULL |
| C | 02/01/19 | NULL |
| A | 02/01/19 | 01/01/19 |
| A | 03/01/19 | 01/01/19 |
| A | 03/01/19 | 02/01/19 |
| B | 03/01/19 | 01/01/19 |
| A | 04/01/19 | 01/01/19 |
| A | 04/01/19 | 02/01/19 |
| A | 04/01/19 | 03/01/19 |
| C | 04/01/19 | 02/01/19 |
| B | 05/01/19 | 01/01/19 |
| B | 05/01/19 | 03/01/19 |
I'm certain this has been answered elsewhere before, but the biggest problem I'm having is not knowing how to word my problem!
EDIT: Really helpful responses so far, but is there a solution where I can change the date "cut-off" eg:
Cut off: 03/01/19
Table: The same
Desired result:
| ID | Sign-in Date | Previous |
| A | 03/01/19 | 02/01/19 |
| B | 03/01/19 | 01/01/19 |
| A | 04/01/19 | 03/01/19 |
| C | 04/01/19 | 02/01/19 |
| B | 05/01/19 | 03/01/19 |

Answer 1

I think that if you need to do that it's better to make an ordering column like:
SELECT *, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY SignInDate) AS O FROM [SIGNIN]
So the end result would be like:
SELECT t.ID, t.SignInDate [Sign-In Date], t2.SignInDate as Previous
FROM (SELECT *, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY SignInDate) AS O FROM [SIGNIN]) t
LEFT JOIN
(SELECT *, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY SignInDate) AS O FROM [SIGNIN]) t2 ON t.ID = t2.ID AND t.O = t2.O+1
Which should give Something akin to:
A 2019-01-01 NULL
A 2019-01-04 2019-01-01
A 2019-02-01 2019-01-04
B 2019-01-01 NULL
B 2019-01-05 2019-01-01
C 2019-01-01 NULL
Hope this helps.

Answer 2

Try using LAG assuming you're on a modern version of SQL Server.
SELECT [SIGNIN].ID,
[SIGNIN].SignInDate,
LAG([SIGNIN].SignInDate) OVER (PARTITION BY [SIGNIN].ID ORDER BY [SIGNIN].SignInDate DESC) AS Previous
FROM [SIGNIN]

Answer 3

using this:
SELECT [SIGNIN].ID,
[SIGNIN].SignInDate,
MAX([Prev].SignInDate) as Previous
FROM [SIGNIN]
LEFT JOIN [SIGNIN] as [Prev] on [SIGNIN].ID = [Prev].ID
and [SIGNIN].SignInDate > [Prev].SignInDate
GROUP BY [SIGNIN].ID, [SIGNIN].SignInDate
ORDER BY [SIGNIN].ID, [SIGNIN].SignInDate

Answer 4

Try something like this:
SELECT
ID, SignInDate,
LAG(SignInDate, 1,SignInDate) OVER(order by ID partition by ID)
FROM SIGNIN

Answer 5

The following will give you almost what you are looking for, just without the nulls.
You should probably do a left outer or right outer join in the inner query, and some extra maneuver to add the null rows as well. I am a lit
select id, max(prev) as prev, signindate from
(
SELECT SIGNIN.ID,
SIGNIN.SignInDate as prev,
prev.signindate
FROM SIGNIN
JOIN SIGNIN as Prev on SIGNIN.ID = Prev.ID
and SIGNIN.SignInDate < Prev.SignInDate
ORDER BY SIGNIN.ID, SIGNIN.SignInDate
) a
group by 1,3

Answer 6

I like the APPLY solution because you can add any amount of columns from the matching row(s):
DECLARE #CutOffDate DATE = '2019-01-03'
SELECT
S.ID,
S.SignInDate,
PreviousSignInDate = R.SignInDate
FROM
[SIGNIN] AS S
OUTER APPLY (
SELECT TOP 1
P.* -- Can incorporate many columns (will also have to add them on the outmost SELECT list)
FROM
SIGNIN AS P
WHERE
S.ID = P.ID AND
P.SignInDate < S.SignInDate
ORDER BY
P.SignInDate DESC
) AS R
WHERE
S.SignInDate >= #CutOffDate
ORDER BY
S.SignInDate,
S.ID
For this case, you can use TOP 1 + ORDER BY to fetch the previous one, as long as you have the link S.ID = P.ID and making sure that P.SignInDate < S.SignInDate.
Also get used to writing dates on the YYYY-MM-DD format, since 03/01/19 might lead to confusions.

Answer 7

A correlated subquery is a very simple solution :
SELECT ID, SignInDate,
(SELECT top 1 SigInDate
FROM SIGNIN as S2
WHERE S2.ID = S1.ID and S2.SignInDate < S1.SignInDate
ORDER BY S2.SignInDate desc) as Previous
FROM SIGNIN as S1
ORDER BY S1.ID, S1.SignInDate