*** WRONG: Because there is no ‘Updated’ virtual table.

2) UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Updated.Hourly as OldPay, Deleted.Hourly as NewPay WHERE Hourly IS NOT NULL;

*** WRONG: Because there is no ‘Updated’ virtual table.

3) UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Inserted.Hourly as OldPay, Deleted.Hourly as NewPay WHERE Hourly IS NOT NULL;

*** WRONG: Because Deleted.Hourly Should be the OLD value not NEW value.

4) UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay WHERE Hourly IS NOT NULL;

**** CORRECT answer.

cause update use two temporary table inserted and Deleted . In deleted it store old data and in inserted it store the new data .

4.UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

You have an HourlyPay table and are giving all hourly employees a $1 raise. When you run the update statement you want to see the EmpID, OldPay, and NewPay. What code will achieve this result?

Answer is 4.
UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

following is also Correct But Alias names are Wrong
3. UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Inserted.Hourly as OldPay, Deleted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Following are Wrong
1. UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Updated.Hourly as NewPay
WHERE Hourly IS NOT NULL

2. UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Updated.Hourly as OldPay, Deleted.Hourly as NewPay
WHERE Hourly IS NOT NULL

I want to insert the output values into another #result table , how can it possible.
or i want to assign the output values to a variable?

Please give me the query.

Example:

create table #tbl (id int identity(1,1) , name varchar(10) , dept varchar(10) )
insert into #tbl (name , dept )
output inserted.*
select ‘abhIShek Online4All’ , ‘d’
/*———————————————————-
Result:
id name dept
1 abhIShek Online4All d
———————————————————-
*/

———————————————————-
–i want to insert above result into #result table.
———————————————————-

Thanks in advance,

Regards,
abhIShek Online4all

4.UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

because deleted will give you old record and inserted will give you new record…

India

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Eric
USA

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

- Somnath Desai
India

# UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

From India

4.UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Country : India

A.) Since Updated is not one of the temporary tables created by the UPDATE statement (1) and (2) are both incorrect answers. Because Inserted.Hourly will contain the new value that was just inserted by the UPDATE statement and Deleted.Hourly will contain the old value that was just deleted (3) will not be correct either. Since the code in (4) is referring to the Deleted.Hourly field as the OldPay and the Inserted.Hourly field as the NewPay it is the correct answer.

Winner from USA: Bill Pepping

Winner from India: Gopalakrishnan Arthanarisamy

I thank you all for participating here. The permanent record of this update is posted on facebook page.

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Rajesh Garg
India

Scalable- Systems

Manoj Sahoo

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

as the deletion of records happens first before insertion we must look for oldpay in deleted table and newpay in insertion table.

USA

Here we want, empId which we are updating, so we can take this from inserted table or deleted table.
Old pay, first we delete old pay, we get this data from deleted table.
New pay,next we insert with new pay , we get this from inserted table.

bhasker
USA

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Options 1 and 2 reference a non-existent table (the “UPDATE” table). Option 3 is misleading as it would return the inverted values for OldPay and NewPay. Option 4 is correct.

Andre Soliz
USA

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL

Location : India

UPDATE HourlyPay SET Hourly = Hourly + 1
OUTPUT Deleted.EmpID , Deleted.Hourly as OldPay, Inserted.Hourly as NewPay
WHERE Hourly IS NOT NULL