Solved

Update Trigger - multiple fields

Posted on 2004-08-19
10
258 Views
Last Modified: 2006-11-17
Greetings Experts...I'm trying to update 2 fields in Table2 when the same fields for the same ID have been updated in Table1. This may be simple, but I have a deadline that is FAST approaching. Can somebody please help me resolve the following statement/error:
----------------------------------------------------------------------------------
CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted.deleted
WHERE Table2.FieldA = inserted.FieldA     'FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
go
-------------------------------------------------------------------------------------
Server: Msg 170, Level 15, State 1, Procedure Update_Rig_Loc, Line 9
Line 9: Incorrect syntax near 'FieldA'.
0
Comment
Question by:rsanglim
  • 3
  • 3
  • 2
  • +1
10 Comments
 
LVL 75

Expert Comment

by:Anthony Perkins
ID: 11846577
CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted, table2
WHERE Table2.FieldA = inserted.FieldA     'FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
go
0
 
LVL 17

Expert Comment

by:BillAn1
ID: 11846582
Nearly there ...... try this

CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted.deleted
WHERE Table2.FieldA = inserted.FieldA  
END  
0
 
LVL 69

Expert Comment

by:ScottPletcher
ID: 11846584
Comments are preceded by -- not ', and you need an 'end' for the 'begin', so try this:


CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted
WHERE Table2.FieldA = inserted.FieldA     --FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
END
go
0
 
LVL 75

Expert Comment

by:Anthony Perkins
ID: 11846585
Or:
CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted
          Inner Join Table2 On inserted.FieldA = Table2.FieldA  'FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
go
0
 
LVL 17

Accepted Solution

by:
BillAn1 earned 500 total points
ID: 11846592
Oops - didn't spot the 'deleted' word

Nearly there ...... try this

CREATE TRIGGER Update_FieldB_FieldC ON Table1 FOR UPDATE AS

IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM inserted
WHERE Table2.FieldA = inserted.FieldA  
END  
0
PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

 
LVL 75

Expert Comment

by:Anthony Perkins
ID: 11846597
Scott as usual :) has it right.
0
 
LVL 69

Expert Comment

by:ScottPletcher
ID: 11846599
Or, perhaps very slightly clearer:


CREATE TRIGGER Update_FieldB_FieldC
ON Table1
FOR UPDATE AS
IF UPDATE (FieldB) or UPDATE (FieldC)
BEGIN
UPDATE Table2
SET Table2.FieldB = inserted.FieldB ,
Table2.FieldC = inserted.FieldC
FROM Table2
INNER JOIN inserted ON Table2.FieldA = inserted.FieldA  --FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
END
go
0
 

Author Comment

by:rsanglim
ID: 11846701
The comments aren't part of the statement. I included to try to be more descriptive. Sorry for any confusion, and I'll properly notate in the future (my Access background, you know). Working the solutions now...
0
 

Author Comment

by:rsanglim
ID: 11846831
pts to BillAn1 - works great!

The ScottPletcher / acperkins solutions produced the following:

Server: Msg 207, Level 16, State 3, Procedure Update_Rig_Loc, Line 4
Invalid column name Table2.

acperkins first solution produced original error.

Thanks, everyone, for the help...REALLY appreciate it.
0
 
LVL 69

Expert Comment

by:ScottPletcher
ID: 11851467
I suspect you added a carriage return/line feed on this line:

INNER JOIN inserted ON Table2.FieldA = inserted.FieldA  --FieldA is ID in Table1, inserted to {CR?} Table2 via trigger (not ID in Table2)


That was a *single* line, not two.  It might be clearer to see like this:

--FieldA is ID in Table1, inserted to Table2 via trigger (not ID in Table2)
INNER JOIN inserted ON Table2.FieldA = inserted.FieldA  
0

Featured Post

PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Why is this different from all of the other step by step guides?  Because I make a living as a DBA and not as a writer and I lived through this experience. Defining the name: When I talk to people they say different names on this subject stuff l…
Slowly Changing Dimension Transformation component in data task flow is very useful for us to manage and control how data changes in SSIS.
This video shows, step by step, how to configure Oracle Heterogeneous Services via the Generic Gateway Agent in order to make a connection from an Oracle session and access a remote SQL Server database table.
Using examples as well as descriptions, and references to Books Online, show the documentation available for datatypes, explain the available data types and show how data can be passed into and out of variables.

930 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

13 Experts available now in Live!

Get 1:1 Help Now