Check duplicates in sql table and replace the duplicates ID in another table - sql

Question

I have a table with duplicate entries (I forgot to make NAME column unique)
So I now have this Duplicate entry table called 'table 1'
ID NAME
1 John F Smith
2 Sam G Davies
3 Tom W Mack
4 Bob W E Jone
5 Tom W Mack
IE ID 3 and 5 are duplicates
Table 2
ID NAMEID ORDERS
1 2 item4
2 1 item5
3 4 item6
4 3 item23
5 5 item34
NAMEID are ID from table 1. Table 2 ID 4 and 5 I want to have NAMEID of 3 (Tom W Mack's Orders) like so
Table 2 (correct version)
ID NAMEID ORDERS
1 2 item4
2 1 item5
3 4 item6
4 3 item23
5 3 item34
Is there an easy way to find and update the duplicates NAMEID in table 2 then remove the duplicates from table 1

Answer 1

In this case what you can do is.
You can find how many duplicate records you have.
In Order to find duplicate records you can use.
SELECT ID, NAME,COUNT(1) as CNT FROM TABLE1 GROUP BY ID, NAME
This is will give you the count and you find all the duplicate records
and delete them manually.
Don't forget to alter your table after removing all the duplicate records.

Answer 2

Here's how you can do it:
-- set up the environment
create table #t (ID int, NAME varchar(50))
insert #t values
(1, 'John F Smith'),
(2, 'Sam G Davies'),
(3, 'Tom W Mack'),
(4, 'Bob W E Jone'),
(5, 'Tom W Mack')
create table #t2 (ID int, NAMEID int, ORDERS varchar(10))
insert #t2 values
(1, 2, 'item4'),
(2, 1, 'item5'),
(3, 4, 'item6'),
(4, 3, 'item23'),
(5, 5, 'item34')
go
-- update the referencing table first
;with x as (
select id,
first_value(id) over(partition by name order by id) replace_with
from #t
),
y as (
select #t2.nameid, x.replace_with
FROM #t2
join x on #t2.nameid = x.id
where #t2.nameid <> x.replace_with
)
update y set nameid = replace_with
-- delete duplicates from referenced table
;with x as (
select *, row_number() over(partition by name order by id) rn
from #t
)
delete x where rn > 1
select * from #t
select * from #t2
Pls, test first for performance and validity.

Answer 3

Let's use the example data
INSERT INTO TableA
(`ID`, `NAME`)
VALUES
(1, 'NameA'),
(2, 'NameB'),
(3, 'NameA'),
(4, 'NameC'),
(5, 'NameB'),
(6, 'NameD')
and
INSERT INTO TableB
(`ID`, `NAMEID`, `ORDERS`)
VALUES
(1, 2, 'itemB1'),
(2, 1, 'itemA1'),
(3, 4, 'itemC1'),
(4, 3, 'itemA2'),
(5, 5, 'itemB2'),
(5, 6, 'itemD1')
(makes it a bit easier to spot the duplicates and check the result)
Let's start with a simple query to get the smallest ID for a given NAME
SELECT
NAME, min(ID)
FROM
tableA
GROUP BY
NAME
And the result is [NameA,1], [NameB,2], [NameC,4], [NameD,6]
Now if you use that as an uncorrelated subquery for a JOIN with the base table like
SELECT
keep.kid, dup.id
FROM
tableA as dup
JOIN
(
SELECT
NAME, min(ID) as kid
FROM
tableA
GROUP BY
NAME
) as keep
ON
keep.NAME=dup.NAME
AND keep.kid<dup.id
It finds all duplicates that have the same name as in the result of the subquery but a different id + it also gives you the id of the "original", i.e. the smallest id for that name.
For the example it's [1,3], [2,5]
Now you can use that in an UPDATE query like
UPDATE
TableB as b
JOIN
tableA as dup
JOIN
(
SELECT
NAME, min(ID) as kid
FROM
tableA
GROUP BY
NAME
) as keep
ON
keep.NAME=dup.NAME
AND keep.kid<dup.id
SET
b.NAMEID=keep.kid
WHERE
b.NAMEID=dup.id
And the result is
ID,NAMEID,ORDERS
1, 2, itemB1
2, 1, itemA1
3, 4, itemC1
4, 1, itemA2 <- now has NAMEID=1
5, 2, itemB2 <- now has NAMEID=2
5, 6, itemD1
To eleminate the duplicates from tableA you can use the first query again.