我有一张像这样的桌子:

编号

数字=整数,编码=整数

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$$


拨打create(21377,54,1);

始终给我变量@Exist为1,因此即使数字和鳕鱼的组合不存在,也不要继续进行。

谁能指出我做错了什么?
谢谢。

最佳答案

尝试使用select into代替set xx =(select ...)。

还有cod = Cod总是会返回true ...

您还需要声明变量...

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$$

08-17 22:54