MySQL Bugs: #35430: "if" example from documentation doesn't work

Bug #35430	"if" example from documentation doesn't work
Submitted:	19 Mar 2008 9:52	Modified:	19 Mar 2008 21:14
Reporter:	Daniel Fischer	Email Updates:
Status:	Closed	Impact on me:	None
Category:	MySQL Server: Documentation	Severity:	S3 (Non-critical)
Version:	5.0+	OS:	Any
Assigned to:	Jon Stephens	CPU Architecture:	Any

Description:
http://dev.mysql.com/doc/refman/5.0/en/if-statement.html

DELIMITER //

CREATE FUNCTION SimpleCompare(n INT, m INT) RETURNS VARCHAR(20)
  BEGIN
    DECLARE s VARCHAR(20);

    IF n > m THEN s = '>';
    ELSEIF n = m THEN s = '=';
    ELSE s = '<'
    END IF;

    s = CONCAT(n, ' ', s, ' ', m);

    RETURN s;
  END //

DELIMITER ;

produces this error:
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 '= '>';

How to repeat:
Just paste the code into the command line client.

Suggested fix:
DELIMITER //

CREATE FUNCTION SimpleCompare(n INT, m INT) RETURNS VARCHAR(20)
  BEGIN
    DECLARE s VARCHAR(20);

    IF n > m THEN SET s = '>';
    ELSEIF n = m THEN SET s = '=';
    ELSE SET s = '<'; 
    END IF;

    SET s = CONCAT(n, ' ', s, ' ', m);

    RETURN s;
  END //

DELIMITER ;

Thank you for a bug report.

This is my goof - I'll handle it.

Thank you for your bug report. This issue has been addressed in the documentation. The updated documentation will appear on our website shortly, and will be included in the next release of the relevant products.