Create a function for generating random number in SQL Server trigger - sql

Question

I have to create a function in a SQL Server trigger for generating random numbers after insert. I want to update the column with that generated random number please help what I have missed in my code.
If you know other ways please suggest a way to complete my task.
This my SQL Server trigger:
ALTER TRIGGER [dbo].[trgEnquiryMaster]
ON [dbo].[enquiry_master]
AFTER INSERT
AS
declare #EnquiryId int;
declare #ReferenceNo varchar(50);
declare #GenReferenceNo NVARCHAR(MAX);
select #EnquiryId = i.enquiry_id from inserted i;
select #ReferenceNo = i.reference_no from inserted i;
BEGIN
SET #GenReferenceNo = 'CREATE FUNCTION functionRandom (#Reference VARCHAR(MAX) )
RETURNS VARCHAR(MAX)
As
Begin
DECLARE #r varchar(8);
SELECT #r = coalesce(#r, '') + n
FROM (SELECT top 8
CHAR(number) n FROM
master..spt_values
WHERE type = P AND
(number between ascii(0) and ascii(9)
or number between ascii(A) and ascii(Z)
or number between ascii(a) and ascii(z))
ORDER BY newid()) a
RETURNS #r
END
'
EXEC(#GenReferenceNo)
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
SET NOCOUNT ON
-- update statements for trigger here
UPDATE enquiry_master
SET reference_no ='updated'
WHERE enquiry_id = #EnquiryId
END

Answer 1

To generate random numbers, just call CRYPT_GEN_RANDOM which was introduced in SQL Server 2008:
SELECT CRYPT_GEN_RANDOM(5) AS [Hex],
CONVERT(VARCHAR(20), CRYPT_GEN_RANDOM(5), 2) AS [HexStringWithout0x],
CONVERT(VARCHAR(20), CRYPT_GEN_RANDOM(10)) AS [Translated-ASCII],
CONVERT(NVARCHAR(20), CRYPT_GEN_RANDOM(20)) AS [Translated-UCS2orUTF16]
returns:
Hex HexStringWithout0x Translated-ASCII Translated-UCS2orUTF16
0x4F7D9ABBC4 0ECF378A7A ¿"bü<ݱØï 붻槬㟰添䛺⯣왚꒣찭퓚
If you are ok with just 0 - 9 and A - F, then the CONVERT(VARCHAR(20), CRYPT_GEN_RANDOM(5), 2) is all you need.
Please see my answer on DBA.StackExchange on a similar question for more details:
Password generator function
The UPDATE statement shown in the "Update" section of that linked answer is what you want, just remove the WHERE condition and add the JOIN to the Inserted pseudo-table.
The query should look something like the following:
DECLARE #Length INT = 10;
UPDATE em
SET em.[reference_no] = rnd.RandomValue
FROM dbo.enquiry_master em
INNER JOIN Inserted ins
ON ins.enquiry_id = em.enquiry_id
CROSS APPLY dbo.GenerateReferenceNo(CRYPT_GEN_RANDOM((em.[enquiry_id] % 1) + #Length)) rnd;
And since the function is slightly different, here is how it should be in order to get both upper-case and lower-case letters:
CREATE FUNCTION dbo.GenerateReferenceNo(#RandomValue VARBINARY(20))
RETURNS TABLE
WITH SCHEMABINDING
AS RETURN
WITH base(item) AS
(
SELECT NULL UNION ALL SELECT NULL UNION ALL SELECT NULL UNION ALL
SELECT NULL UNION ALL SELECT NULL UNION ALL SELECT NULL
), items(item) AS
(
SELECT NULL
FROM base b1
CROSS JOIN base b2
)
SELECT (
SELECT TOP (LEN(#RandomValue))
SUBSTRING('1234567890QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm',
(CONVERT(TINYINT, SUBSTRING(#RandomValue, 1, 1)) % 62) + 1,
1) AS [text()]
FROM items
FOR XML PATH('')
) AS [RandomReferenceNo];
GO
And please follow the usage shown above, passing in CRYPT_GEN_RANDOM((em.[enquiry_id] % 1) + #Length), not: CRYPT_GEN_RANDOM(#RefferenceNOLength).
Other notes:
#marc_s already explained the one-row vs multiple-rows flaw and how to fix that.
not only is a trigger not the place to create a new object (i.e. the function), that function wouldn't have worked anyway since the call to newid() (in the ORDER BY) is not allowed in a function.
You don't need to issue two separate SELECTs to set two different variables. You could do the following:
SELECT #EnquiryId = i.enquiry_id,
#ReferenceNo = i.reference_no
FROM TableName i;
Passing strings into a function requires quoting those strings inside of single-quotes: ASCII('A') instead of ASCII(A).
UPDATE
The full Trigger definition should be something like the following:
ALTER TRIGGER [dbo].[trgEnquiryMaster]
ON [dbo].[enquiry_master]
AFTER INSERT
AS
BEGIN
DECLARE #Length INT = 10;
UPDATE em
SET em.[reference_no] = rnd.RandomValue
FROM dbo.enquiry_master em
INNER JOIN Inserted ins
ON ins.enquiry_id = em.enquiry_id
CROSS APPLY dbo.GenerateReferenceNo(
CRYPT_GEN_RANDOM((em.[enquiry_id] % 1) + #Length)
) rnd;
END;

Answer 2

A trigger should be very nimble and quick - it is no place to do heavy and time-intensive processing, and definitely no place to create new database objects since (a) the trigger is executed in the context of the code causing it to fire, and (b) you cannot control when and how often the trigger is fired.
You need to
define and create your function to generate that random value during database setup - once, before any operations are executed on the database
rewrite your trigger to take into account that multiple rows could be inserted at once, and in that case, the Inserted table will contain multiple rows which all have to be handled.
So your trigger will look something like this (with several assumptions by me - e.g. that enquiry_id is the primary key on your table - you need this to establish the INNER JOIN between your data table and the Inserted pseudo table:
ALTER TRIGGER [dbo].[trgEnquiryMaster]
ON [dbo].[enquiry_master]
AFTER INSERT
AS
-- SET NOCOUNT ON added to prevent extra result sets from
-- interfering with SELECT statements.
SET NOCOUNT ON
-- update statements for trigger here
UPDATE enq
SET reference_no = dbo.GenerateRandomValue(.....)
FROM enquiry_master enq
INNER JOIN inserted i ON enq.enquiry_id = i.enquiry_id