MySQL stored procedure return value

Question

I have to create an SP that returns a value if it\'s valid or not. But it doesn\'t return anything and I don\'t know, why?

CREATE DEFINER=`root`@`localhost`          


        

Answer 1

Add:

• DELIMITER at the beginning and end of the SP.
• DROP PROCEDURE IF EXISTS validar_egreso; at the beginning
• When calling the SP, use @variableName.

This works for me. (I modified some part of your script so ANYONE can run it with out having your tables).

DROP PROCEDURE IF EXISTS `validar_egreso`;

DELIMITER $$

CREATE DEFINER='root'@'localhost' PROCEDURE `validar_egreso` (
    IN codigo_producto VARCHAR(100),
    IN cantidad INT,
    OUT valido INT(11)
)
BEGIN

    DECLARE resta INT;
    SET resta = 0;

    SELECT (codigo_producto - cantidad) INTO resta;

    IF(resta > 1) THEN
       SET valido = 1;
    ELSE
       SET valido = -1;
    END IF;

    SELECT valido;
END $$

DELIMITER ;

-- execute the stored procedure
CALL validar_egreso(4, 1, @val);

-- display the result
select @val;

Answer 2

Update your SP and handle exception in it using declare handler with get diagnostics so that you will know if there is an exception. e.g.

CREATE DEFINER=`root`@`localhost` PROCEDURE `validar_egreso`(
IN codigo_producto VARCHAR(100),
IN cantidad INT,
OUT valido INT(11)
)
BEGIN
DECLARE EXIT HANDLER FOR SQLEXCEPTION
BEGIN
    GET DIAGNOSTICS CONDITION 1
    @p1 = RETURNED_SQLSTATE, @p2 = MESSAGE_TEXT;
    SELECT @p1, @p2;
END
DECLARE resta INT(11);
SET resta = 0;

SELECT (s.stock - cantidad) INTO resta
FROM stock AS s
WHERE codigo_producto = s.codigo;

IF (resta > s.stock_minimo) THEN
    SET valido = 1;
ELSE
    SET valido = -1;
END IF;
SELECT valido;
END

Answer 3

You have done the stored procedure correctly but I think you have not referenced the valido variable properly. I was looking at some examples and they have put an @ symbol before the parameter like this @Valido

This statement SELECT valido; should be like this SELECT @valido;

Look at this link mysql stored-procedure: out parameter. Notice the solution with 7 upvotes. He has reference the parameter with an @ sign, hence I suggested you add an @ sign before your parameter valido

I hope that works for you. if it does vote up and mark it as the answer. If not, tell me.