Generate random records for table in SQL I'm doing a project in ms sql and I have to fill table with random records.

Here is a definition of table:

create table invoice(
    id_invoice INT PRIMARY KEY IDENTITY(1,1),
    id_employee INT FOREIGN KEY REFERENCES Employee(id_employee),
    id_client INT FOREIGN KEY REFERENCES Client(id_client),
    id_service INT FOREIGN KEY REFERENCES Service(id_service),
    amount MONEY,
    payment_way VARCHAR(20) CHECK (payment_way in ('cash', 'credit_card'))
);


And here is the code that generates random records:

ALTER PROCEDURE generate_invoices
AS
BEGIN
    DECLARE @id_employee INT
    DECLARE @id_client INT
    DECLARE @id_service INT
    DECLARE @amount INT
    DECLARE @payment_way VARCHAR(20)

    SET @id_employee = (SELECT TOP 1 id_employee FROM Employee ORDER BY NEWID())
    SET @id_client = (SELECT TOP 1 id_client FROM Client ORDER BY NEWID())
    SET @id_service = (SELECT TOP 1 id_service FROM Service ORDER BY NEWID())
    EXEC random 1000, 5000, @amount OUTPUT
    IF (@amount % 2 = 1) SET @payment_way = 'cash'
    ELSE SET @payment_way = 'credit card'
    INSERT INTO Invoice VALUES (id_employee, @id_clienta, @id_service, @amount, @payment_way)
END
GO

EXEC generate_invoices
GO 10


The code works obviously, but I would like to ask if there is better way (shorter, more efficient) to do same thing, because I must write over a dozen of similar procedure for other tables.
Randomness
Using NEWID() seems like a pretty smart way of getting random records, and it is. There is however a better function for this purpose. Say hello to the CRYPT_GEN_RANDOM() function.
The CRYPT_GEN_RANDOM() has a parameter that we need to provide, the length of the random string it is going to generate. For what we need, 4 or 5 characters is plenty.
I will update my answer with the proof when I find the time, but my own testing found CRYPT_GEN_RANDOM() generates less predictable records than NEWID().
The Random Stored Procedure
I have no idea what your Random stored procedure entails, but might I suggest an alternate method of generating a random number...
WITH Numbers(I) AS
(
    SELECT 1000
    
    UNION ALL 
    
    SELECT I+1
    FROM Numbers
    WHERE I&lt;5000
)

SELECT TOP 1 @amount = I
FROM Numbers
ORDER BY CRYPT_GEN_RANDOM(4)
OPTION (MAXRECURSION 0)

First, this generates a table of numbers 1000 to 5000 using a recursive CTE, then selects one randomly using the CRYPT_GEN_RANDOM() that we discussed earlier.
To If or Not to If
Might I also suggest an alternate method of generating the @payment_way. Instead of using an If, might I suggest using a Simple Case Statement.
Personally I find this
SET @payment_way = 
CASE @amount % 2
    WHEN 1
    THEN 'cash'
    ELSE 'credit card'
END

To be a more SQL friendly way of writing it.
Putting it all together
This is what I ended up with. It's not any smaller, but it doesn't rely on any external stored procedures.
ALTER PROCEDURE generate_invoices
AS
BEGIN
    DECLARE 
        @id_employee INT = (SELECT TOP 1 id_employee FROM Employee ORDER BY CRYPT_GEN_RANDOM(4)),
        @id_client INT = (SELECT TOP 1 id_client FROM Client ORDER BY CRYPT_GEN_RANDOM(4)),
        @id_service INT = (SELECT TOP 1 id_service FROM Service ORDER BY CRYPT_GEN_RANDOM(4)),
        @amount INT,
        @payment_way VARCHAR(20)

    SET @id_employee 
    SET @id_client 
    SET @id_service 

    WITH Numbers(I) AS
    (
        SELECT 1000
        
        UNION ALL 
        
        SELECT I+1
        FROM Numbers
        WHERE I&lt;5000
    )
    SELECT TOP 1 @amount = I
    FROM Numbers
    ORDER BY CRYPT_GEN_RANDOM(4)
    OPTION (MAXRECURSION 0)

    SET @payment_way = 
    CASE @amount % 2
        WHEN 1
        THEN 'cash'
        ELSE 'credit card'
    END

    INSERT INTO Invoice VALUES (@id_employee, @id_clienta, @id_service, @amount, @payment_way)
END
GO