Update SQL table with random value from other table

Question

On Microsoft SQL Server 2008, I have a table with Products:

Id | Name | DefaultImageId

And one with Images:

Id | ProductId | Bytes

I want to run an Update statement, that updates the DefaultImageId on all records in the Products table with a random Id from the Images table that is related to the Product via the ProductId column.

Can anyone help out? Should be simple for any SQL Champ (Which is obviously not me)..

score 23 · Answer 1 · answered Sep 22 '15 at 11:56

23

Addressing @philreed's issue with the selected answer:

Is there a way to assign each row being updated with a different value randomly chosen from the source table?

UPDATE Products
SET DefaultImageId = t2.Id
FROM Products t1
CROSS APPLY (
    SELECT TOP 1 Id
    FROM Images
    WHERE t1.Id = t1.Id
    ORDER BY newid()
    ) t2

answered Sep 22 '15 at 11:56

jacoblambert

787
1
11
18

5

This solved my issue with getting the same random number on each row. The "where t1.Id = t1.Id" is needed to make it work, I initially didn't notice it and though the cross apply was the only difference between the accepted answer and this one. – Luke Rice Dec 09 '17 at 16:18
1

For the record (no pun intended), this is not very performant if used against a large-ish dataset with (I'm updating a phone number from a list of 10000 random phone numbers to scramble a table of 300k members and am currently at 33min and wondering how much longer...) – jleach Feb 15 '18 at 16:06
maybe you could post with details? – jacoblambert Feb 15 '18 at 16:26
1

@KrishnanVenkiteswaran This is not the correct answer as it randomly assigns any Image.Id to the Product.DefaultImageId table, not only those that are related to the Product. – Robin Day Jan 11 '19 at 09:27
I was looking for a random update of the first table with values from the 2nd table and Cross Apply was what worked for me, OP's problem is a different one though. – Krishnan Venkiteswaran Jan 11 '19 at 11:04

Robin Day · Accepted Answer · 2019-01-11T09:22:45.157

You can do an order by on a NEWID to get a random number for every row of your update.

UPDATE
    Products
SET
    DefaultImageId =
    (
        SELECT TOP 1
            Id
        FROM
            Images
        WHERE
            Images.ProductId = Products.Id
        ORDER BY
            NEWID()
    )

This has been down marked and comments added indicating it does not solve the problem. I think the confusion has come from where people have not realised the original question requests a random image be selected for each product, hence the where clause with Product Id. Have provided a full script with data set below. It adds five products and three images for each product. Then randomly sets the default image id for each product.

CREATE TABLE Products(Id INT, Name NVARCHAR(100), DefaultImageId INT NULL)

CREATE TABLE Images (Id INT, ProductId INT, Bytes VARBINARY(100))

INSERT INTO Products (Id, NAME, DefaultImageId) VALUES(1, 'A', NULL)
INSERT INTO Products (Id, NAME, DefaultImageId) VALUES(2, 'B', NULL)
INSERT INTO Products (Id, NAME, DefaultImageId) VALUES(3, 'C', NULL)
INSERT INTO Products (Id, NAME, DefaultImageId) VALUES(4, 'D', NULL)
INSERT INTO Products (Id, NAME, DefaultImageId) VALUES(5, 'E', NULL)

INSERT INTO Images (Id, ProductId, Bytes) VALUES(1, 1, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(2, 1, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(3, 1, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(4, 2, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(5, 2, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(6, 2, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(7, 3, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(8, 3, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(9, 3, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(10, 4, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(11, 4, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(12, 4, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(13, 5, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(14, 5, NULL)
INSERT INTO Images (Id, ProductId, Bytes) VALUES(15, 5, NULL)

UPDATE
    Products
SET
    DefaultImageId =
    (
        SELECT TOP 1
            Id
        FROM
            Images
        WHERE
            Images.ProductId = Products.Id
        ORDER BY
            NEWID()
    )

SELECT * FROM Products

This solution appears to assign the same randomly chosen value to each row being updated. Is there a way to assign each row being updated with a different value randomly chosen from the source table (Images in your example)? — philreed, Jul 02 '15 at 13:43
Please see if the solution just posted solves this, @philreed — jacoblambert, Sep 22 '15 at 11:58
@KrishnanVenkiteswaran in what way isn't it working? I have updated the answer with create table and data inserts to prove it provides the solution. — Robin Day, Jan 11 '19 at 09:23
Yes, it does work, apologies, I didn't read the question, I was looking at just assigning random values from table to another. — Krishnan Venkiteswaran, Jan 11 '19 at 10:49

score 5 · Answer 3 · answered Oct 11 '13 at 05:48

5

Another possible solution

UPDATE
    Products
SET
    DefaultImageId =
    (
        SELECT TOP 1
            Id
        FROM
            Images
        ORDER BY
            NEWID(), Products.Id
    )

answered Oct 11 '13 at 05:48

Welly

123
1
8

Great this will not save null values if image table length less than product – ASalameh Aug 11 '15 at 17:31

celper · Answer 4 · 2021-07-31T07:10:15.313

Try this one (on AdventureWorks):

It updates all rows of the person table with a random name from the product table.

/*select rows from both tables in random order and match records by row number */
update person.person
set 
FirstName = otherTable.Name
from 
(select BusinessEntityID, row_number() over (order by newid()) as row_num from person.person) tableWithRownum
,(select ProductId, Name, row_number() over (order by newid()) as row_num from production.product) as otherTable
where 
person.person.BusinessEntityID=tableWithRownum.BusinessEntityID 
and (tableWithRownum.row_num%500)=otherTable.row_num /*match by row number;*/
/* optional: correct for different table sizes with modulo operation*/

Notes:

The new values are not sampled independently for each row.
Since the random order is determined only once, all entries occur at roughly the same rate (depending on table sizes: If both tables have the same number of rows, each entry should occur exactly once).
This behaviour might not fit every use case, but it should help performance...

With my testing this has proved to work faster than the cross apply solution.Much much faster. Down from 42 mins to 20 seconds on a slow Azure SQL database. Thank you. — Iain M Norman, Aug 29 '18 at 13:37
Although this is more complex than other answers, it is more versatile and faster. It should be the accepted answer. — Alexei - check Codidact, Jul 28 '21 at 07:52

Raj More · Answer 5 · 2009-08-17T18:45:31.220

1

Check this out.

Update Products
Set DefaultImageId = (
SELECT top 1 Id 
From Images 
Where 1=1
and Products.Id = Images.ProductId
ORDER BY NEWID()
)

edited Aug 17 '09 at 18:45

answered Aug 17 '09 at 18:38

Raj More

47,048
33
131
198

Msg 147, Level 15, State 1, Line 2 An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING clause or a select list, and the column being aggregated is an outer reference. – MartinHN Aug 17 '09 at 18:43

score -1 · Answer 6 · answered Mar 26 '20 at 22:36

declare @carg_id as int
DECLARE cursor_cargo CURSOR FOR 
    SELECT carg_id FROM cargo

OPEN cursor_cargo

FETCH NEXT FROM cursor_cargo
INTO @carg_id
WHILE  @@FETCH_STATUS =0
BEGIN

    update cargo set ciud_id = (
    select top 1 ciud_id from ciudad 
    inner join Pais on Pais.pais_id = ciudad.pais_id
    where ciudad.pais_id is not null and empr_id is  null and pais_status =1 order by  NEWID()) WHERE CARG_ID = @CARG_ID

    FETCH NEXT FROM cursor_cargo
        INTO @carg_id
END
CLOSE cursor_cargo
DEALLOCATE cursor_cargo

Cursors are generally a bad idea, and the columns you entered don't match the OP's — InbetweenWeekends, Mar 27 '20 at 01:46

Update SQL table with random value from other table

6 Answers6

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