我的桌子有点像这样:
rnumber
number = int,cod = int
SELECT * FROM number WHERE number = 21377 and cod = 55; 返回正确的值;
所以我有一个proc调用来插入:
CREATE DEFINER=`root`@`%` PROCEDURE `create`(IN Numb INT, IN Cod INT, IN qt INT)
BEGIN
SET @Cont = 0;
SET @Init = Numb;
WHILE @Cont < qt DO
SET @Exist = (SELECT count(number) FROM rnumber WHERE number = @Init AND cod = Cod LIMIT 1);
IF @Exist = 0 THEN
INSERT INTO (...)
END IF;
SET @Cont = @Cont + 1;
SET @IniT = @Init + 1;
END WHILE;
END$$
致电create
(21377,54,1);
总是给我变量@Exist为1,所以没有继续if,即使数字和cod的组合也不存在。
有谁能指出我做错了什么? 谢谢。
答案 0 :(得分:0)
尝试使用select into而不是set xx =(select ...)。
同时拥有cod = Cod
将始终返回true ...
您还需要申报变量......
试试这个:
CREATE DEFINER=`root`@`%` PROCEDURE `create`(IN p_Numb INT, IN p_Cod INT, IN p_qt INT)
BEGIN
declare v_Exist integer;
declare v_Init integer;
declare v_Exist integer;
SET v_Cont = 0;
SET v_Exist = 0;
SET v_Init = p_Numb;
WHILE v_Cont < p_qt DO
SELECT count(number) into v_Exist FROM rnumber WHERE number = v_Init AND cod = p_Cod LIMIT 1;
IF v_Exist = 0 THEN
INSERT INTO (...)
END IF;
SET v_Cont = v_Cont +1;
SET v_IniT = v_Init + 1;
END WHILE;
END$$