I don't have a 5.0+ server to test with right now but this should work
--- Daevid Vincent <daevid@stripped> wrote:
> I was using SQLYog 5.03 RC1.
>
> vmware ~ # mysql --version
> mysql Ver 14.12 Distrib 5.0.19, for pc-linux-gnu (i686) using
> readline 5.1
>
> But just to sanity check. I ssh'd in and tried this at the mysql
> command
> line utility:
>
> vmware ~ # mysql somedatabase
> Reading table information for completion of table and column names
> You can turn off this feature to get a quicker startup with -A
>
> Welcome to the MySQL monitor. Commands end with ; or \g.
> Your MySQL connection id is 415 to server version: 5.0.19-log
>
> Type 'help;' or '\h' for help. Type '\c' to clear the buffer.
>
> mysql> delimiter //
> mysql> CREATE TRIGGER upd_check BEFORE UPDATE ON starkeys
> -> FOR EACH ROW
> -> BEGIN
> -> IF NEW.skey < 1 THEN
> ->
> Display all 187 possibilities? (y or n)
> -> EW.skey = 1;
You need a SET command here. That's how MySQL's SQL does assignments.
That's also the wrong name, you need the table name of NEW not EW
SET NEW.skey = 1;
> -> ELSEIF NEW.skey > 9 THEN
> ->
> Display all 187 possibilities? (y or n)
> -> EW.skey = 9;
Same comment here:
SET NEW.skey = 9
> -> END IF;
> -> END;//
(Same comment as from other thread.) No ; after the END or END IF
statements
http://dev.mysql.com/doc/refman/5.0/en/begin-end.html
http://dev.mysql.com/doc/refman/5.0/en/flow-control-constructs.html
> delimiter ;
> ERROR 1064 (42000): You have an error in your SQL syntax; check the
> manual
> that corresponds to your MySQL server version for the right syntax to
> use
> near '.skey = 1;
> ELSEIF NEW.skey > 9 THEN
> EW.skey = 9;
> END IF;
> END' at line 5
> mysql> delimiter ;
> mysql>
>
><snip!>
Shawn Green
Database Administrator
Unimin Corporation - Spruce Pine
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