May 12, 2010 at 9:46 pm
Comments posted to this topic are about the item INSTEAD OF trigger and OUTPUT clause
May 12, 2010 at 11:31 pm
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May 13, 2010 at 1:14 am
thank you... however i will take some time to understand this query ... :-D.. still not so perfect.
May 13, 2010 at 2:18 am
Interesting - it never occurred to me that the identity values wouldn't be available in the inserted table!
So how would you get the identity values of the rows you'd just inserted?
May 13, 2010 at 3:35 am
Lost...
Selected wrong option.
May 13, 2010 at 4:49 am
Good Excellent question. Learned quite a bit in the 2nd link (INSTEAD OF trigger).
FYI, in the explanation, it should be the string 'AnotherString' to be replaced with 'Stub', not 'TestString'.
Wayne
Microsoft Certified Master: SQL Server 2008
Author - SQL Server T-SQL Recipes
May 13, 2010 at 4:51 am
Toreador (5/13/2010)
Interesting - it never occurred to me that the identity values wouldn't be available in the inserted table!So how would you get the identity values of the rows you'd just inserted?
Good question, I didn't think about it when I was writing the QOTD.
Here is one of the possible solutions. We can get the identity values inside the trigger using the OUTPUT clause. These values can be passed outside the trigger via a temporary table.
CREATE TABLE TestTable (id INT IDENTITY, string VARCHAR(100))
GO
CREATE TRIGGER TestTrigger ON TestTable
INSTEAD OF INSERT
AS
IF OBJECT_ID('tempdb..#id') IS NOT NULL
INSERT TestTable (string)
OUTPUT inserted.id
INTO #id
SELECT CASE WHEN string = 'TestString' THEN string ELSE 'Stub' END
FROM inserted
ELSE
INSERT TestTable (string)
SELECT CASE WHEN string = 'TestString' THEN string ELSE 'Stub' END
FROM inserted
GO
CREATE TABLE #id (id INT)
SELECT * FROM TestTable
INSERT TestTable (string)
SELECT x.string
FROM
( SELECT 'TestString' AS string
UNION ALL
SELECT 'AnotherString' AS string
) x
SELECT * FROM #id
GO
Maybe it's not the best solution, but it works 🙂
May 13, 2010 at 4:52 am
Toreador (5/13/2010)
Interesting - it never occurred to me that the identity values wouldn't be available in the inserted table!So how would you get the identity values of the rows you'd just inserted?
By not using an "INSTEAD OF" trigger. But I agree with you... you normally don't go investigating the types of triggers on tables when writing code to insert into them.
Wayne
Microsoft Certified Master: SQL Server 2008
Author - SQL Server T-SQL Recipes
May 13, 2010 at 4:55 am
May 13, 2010 at 5:03 am
WayneS (5/13/2010)
in the explanation, it should be the string 'AnotherString' to be replaced with 'Stub', not 'TestString'
You are right, thanks for the correction!
May 13, 2010 at 5:58 am
I selected "Returns an error". It didn't seem right, but I'm not sure why the following code works
INSERT TestTable (string)
OUTPUT inserted.*
Doesn't the select return more columns (ID and String) than the insert list? That is one of my more common errors...
Great question, BTW. I need to learn more about OUTPUT!
May 13, 2010 at 7:15 am
tlewis-993411 (5/13/2010)
I'm not sure why the following code worksINSERT TestTable (string)
OUTPUT inserted.*
Doesn't the select return more columns (ID and String) than the insert list?
When a row is inserted into a table, SQL Server fills all columns in that row. All these columns are contaned in the 'inserted.*' construct. Some of these values are explicit, and some are implicit (NULL, identity, default value, computed value etc).
Here is an example:
create table #t
( id int identity,
a varchar(10),
b varchar(10) default 'hello',
c varchar(10),
computed_column as b + ' ' + c
)
insert #t (c)
output inserted.*
values ('test')
In the QOTD, the construct 'OUTPUT inserted.*' is equal to 'OUTPUT inserted.id, inserted.string'. 'Inserted.string' is an explicit column, while 'inserted.id' is an impilcit column.
May 13, 2010 at 7:46 am
My problem is that I expected that "inserted" would be a complete copy of the inserted row (I haven't used OUTPUT, but triggers certainly work that way). As such, the Inserted.* should return 2 columns, but the insert list only contains one field. See the code below and the resulting error message. This is almost exactly the same construct.
I'll do some research, but it appears that Output is only returning the "explicitly added" columns?
BTW, generically, I'm opposed to * for this very reason!
Declare @Foo Table
(
ID int,
String varchar(100)
)
Insert into @Foo(String)
Select * from
(Select 1 as intval, 'sample1' as string
union all
Select 2 as intval, 'sample2' as string
) as bar
Msg 121, Level 15, State 1, Line 7
The select list for the INSERT statement contains more items than the insert list. The number of SELECT values must match the number of INSERT columns.
May 13, 2010 at 7:47 am
Toreador (5/13/2010)
Interesting - it never occurred to me that the identity values wouldn't be available in the inserted table!So how would you get the identity values of the rows you'd just inserted?
If you are using an instead of trigger, I'm not sure. However, not using an instead of trigger works fine to retrieve the identity values using OUTPUT. I ran the below on SQL Server 2008.
CREATE TABLE Test
(
ID int identity(1,1) NOT NULL PRIMARY KEY,
AnotherColumn varchar(50) NULL
)
GO
CREATE TRIGGER Test_Insert
ON Test
AFTER INSERT
AS
SELECT 'Trigger Executed'
GO
DECLARE @tTableVar table(TestID int NOT NULL,
TestAnotherColumn varchar(50) NULL)
INSERT INTO Test(AnotherColumn)
OUTPUT inserted.ID, inserted.AnotherColumn
INTO @tTableVar(TestID,TestAnotherColumn)
SELECT 'Blah'
UNION ALL
SELECT 'Blah2'
SELECT TestID,TestAnotherColumn FROM @tTableVar
DROP TABLE Test
May 13, 2010 at 7:55 am
Duh - The "Into @tab" is not using the Select, but instead the output, which does include two columns.
Select * into a table (or Output * into table, in this case) only works if the fields are selected in the same order and same count as the target, which is obviously true in this case. However, I would still avoid the * if possible, explicitly naming the fields instead. Otherwise, a change to the target table would break that condition.
Thanks for putting up with me working through my own question. That's how I learn almost everything!
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