Adding Random Id for each unique value in table - sql

Question

I have the table like
ID RANDOM_ID
1 123
10 456
25 789
1 1112
55 1314
10 1516
I want the result to be like :
ID RANDOM_ID
1 123
10 456
25 789
1 123
55 1314
10 456
The same ID should have same random_ids. I'm using the update statement to generate the Random_IDs after creating the table.
CREATE TABLE [RANDOMID_TABLE]([ID] [int] NULL, [RANDOM_ID] [int] NULL)
GO
INSERT INTO [RANDOMID_TABLE] ([ID])
select distinct ABC_ID from RANDOMID_ABC
GO
******** This is the update statement for the RANDOM_ID column in
[RANDOMID_TABLE] table ************
UPDATE [RANDOMID_TABLE]
SET RANDOM_ID = abs(checksum(NewId()) % 1000000)
Is there something else that I need to add to the update statement?
Please advise.

Answer 1

Why would you use update for this? Just generate the values when you insert them:
insert into [RANDOMID_TABLE] (ID, RANDOM_ID)
select ABC_ID, abs(checksum(NewId()) % 1000000)
from RANDOMID_ABC
group by ABC_ID;
EDIT:
If your problem is collisions, then fix how you do the assignment. Just assign a number . . . randomly:
insert into [RANDOMID_TABLE] (ID, RANDOM_ID)
select ABC_ID, row_number() over (order by newid())
from RANDOMID_ABC
group by ABC_ID;
This is guaranteed to not return duplicates.

Answer 2

At a total guess, are you simpling wanting to UPDATE the table so that all the values of a specific ID to have the same value for Random_ID? Like this?
CREATE TABLE YourTable (ID int, Random_ID int);
INSERT INTO YourTable
VALUES(1 ,123),
(10,456),
(25,789),
(1 ,1112),
(55,1314),
(10,1516);
GO
WITH CTE AS(
SELECT ID,
Random_ID,
MIN(Random_ID) OVER (PARTITION BY ID) AS Min_Random_ID
FROM YourTable)
UPDATE CTE
SET Random_ID = Min_Random_ID;
GO
SELECT *
FROM YourTable;
GO
DROP TABLE YourTable;

Answer 3

Here is the script you need with use of temporary table (you need it to persist your random results for each unique ID):
DECLARE #Tbl TABLE (ID INT, RANDOM_ID INT)
INSERT #Tbl (Id) VALUES(1), (10), (25), (1), (55), (10)
SELECT Id, abs(checksum(NewId()) % 1000000) AS Random_Id INTO #distinctData FROM #Tbl GROUP BY Id
SELECT D.* FROM #Tbl T JOIN #distinctData D ON D.ID = T.ID
DROP TABLE #distinctData
Obviously, you don't need the first two rows where I create and initialize data table
Result:
Id Random_Id
1 354317
1 62026
10 532304
10 604768
25 874209
55 718643

Answer 4

You want one random value per ID. So one should think that the following would work:
with ids as
(
select distinct id
from randomid_table
)
, ids_with_rnd as
(
select id, abs(checksum(NewId()) % 1000000) as rnd
from ids
)
update randomid_table
set random_id =
(
select rnd
from ids_with_rnd
where ids_with_rnd.id = randomid_table.id
);
It doesn't however. SQL Server is somewhat buggy here and still creates different numbers for the same ID.
So, your best bet may be: do your update that does create different values (your original update statement). Then correct the data as follows:
update randomid_table
set random_id =
(
select min(random_id)
from randomid_table rt2
where rt2.id = randomid_table.id
);
Demo: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=504236db66fba0f12dc7e407a51451f8