Beginning Microsoft SQL Server 2008 ... - S3 Tech Training

Go ahead and run this once:
DECLARE @Return int;
EXEC @Return = HumanResources.uspEmployeeHireInfo2
@BusinessEntityID = 1,
@JobTitle = ‘His New Title’,
@HireDate = ‘1996-07-01’,
@RateChangeDate = ‘2008-07-31’,
@Rate = 15,
@PayFrequency = 1,
@CurrentFlag = 1;
SELECT @Return;
And everything seems to run fine, but execute it a second time and we get some different results:
Duplicate Rate Change Found
-----------
-2000
(1 row(s) affected)
We tried to insert a second row with the same pay history, but SQL Server wouldn’t allow that. We’ve
used PRINT to supply informative output in case the statement is being executed without the Query
window, and we’re outputting a specific return value that the client can match against a value in a
resource list.
Now, let’s try the same basic test, but use an invalid BusinessEntityID:
DECLARE @Return int;
EXEC @Return = HumanResources.uspEmployeeHireInfo2
@BusinessEntityID = 99999,
@JobTitle = ‘My Invalid Employee’,;
@HireDate = '2008-07-31',
@RateChangeDate = ‘2008-07-31’,
@Rate = 15,
@PayFrequency = 1,
@CurrentFlag = 1;
SELECT @Return;
We get a similar error message and return code, but each is slightly different to allow for the specific
error detected:
BusinessEntityID Not Found
-----------
-1000
(1 row(s) affected)
Chapter 12: Stored Procedures
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