Adding Random Id for each unique value in table

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

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.