尝试创建MySql触发器

CREATE TRIGGER updVisible AFTER UPDATE ON photos
FOR EACH ROW 
BEGIN
    IF NEW.Status = 2 THEN
        UPDATE otherTable SET IsVisible=0 WHERE PID=NEW.PID
    END IF;
END;

但是我得到了错误：

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 ‘END IF’ at
line 6

MySQL版本：5.1.41-community
我究竟做错了什么？

UPD1.这无济于事

DELIMITER //
CREATE TRIGGER updVisible AFTER UPDATE ON photos
FOR EACH ROW 
BEGIN
    IF NEW.Status = 2 THEN
        UPDATE otherTable SET IsVisible=0 WHERE PID=NEW.PID
    END IF
END//
DELIMETER ;

错误：

Error Code: 1064 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 ‘END IF END’ at line 6

我具有root用户访问权限,并使用MySql Workbench 5.2.31 CE

解决方法:

这适用于我的机器！

mysql> DELIMITER //
mysql> CREATE TRIGGER test1 AFTER UPDATE ON test
    -> FOR EACH ROW
    -> BEGIN
    ->     IF NEW.itemId = '2' THEN
    ->         UPDATE test1 SET col1=0 WHERE col2=NEW.`value`;
    ->     END IF;
    -> END//
Query OK, 0 rows affected (0.05 sec)

mysql> DELIMiTER ;
mysql> desc test;
+--------+--------------+------+-----+---------+-------+
| Field  | Type         | Null | Key | Default | Extra |
+--------+--------------+------+-----+---------+-------+
| itemId | varchar(100) | YES  |     | NULL    |       |
| key    | varchar(100) | YES  |     | NULL    |       |
| value  | varchar(100) | YES  |     | NULL    |       |
+--------+--------------+------+-----+---------+-------+
3 rows in set (0.01 sec)

mysql> desc test1;
+-------+--------------+------+-----+---------+-------+
| Field | Type         | Null | Key | Default | Extra |
+-------+--------------+------+-----+---------+-------+
| col1  | varchar(100) | YES  |     | NULL    |       |
| col2  | varchar(100) | YES  |     | NULL    |       |
+-------+--------------+------+-----+---------+-------+
2 rows in set (0.00 sec)

mysql>