MySQL 中的 FUNCTION ... 错误中找不到 RETURN

Question

我浏览了一下，其他解决方案似乎不适合我。当尝试在 MySQL 中创建此函数时，我不断收到“找不到函数返回”错误。知道为什么吗？

CREATE FUNCTION `mydb`.`myfunction`(Name varchar(20))
RETURNS int
LANGUAGE SQL
NOT DETERMINISTIC
SELECT SUM(Transaction.Bought) INTO @Qty FROM Transaction WHERE Transaction.Name = Name;
RETURN @Qty

Answer 1

试试这个

DELIMITER $$
CREATE FUNCTION `myfunction`(`Name` VARCHAR(20) CHARSET utf8) RETURNS INT NOT DETERMINISTIC
READS SQL DATA
MAIN: BEGIN
DECLARE returnVal int;
SELECT SUM(`Transaction`.Bought) INTO returnVal FROM `Transaction` WHERE `Transaction`.Name = Name;
RETURN returnVal;
END MAIN;$$
DELIMITER ;

Answer 2

或者，试试这个，

DELIMITER $$
CREATE FUNCTION `myfunction`(Name varchar(20))
RETURNS int
LANGUAGE SQL
NOT DETERMINISTIC
BEGIN

  DECLARE returnVal int;

  SELECT SUM(`Transaction`.Bought) INTO returnVal 
  FROM `Transaction` 
  WHERE `Transaction`.Name = Name;

  RETURN returnVal;

END$$
DELIMITER ;

Answer 3

delimiter //
create function myfun1234 (i int,j int)
returns int 
begin 
      declare x int ;
       declare    y int ;
        declare  z int ;
      set x=10;
      set y=20;
      if (1<=x && j>=y )then 
      set z=i+j;
      end if;
    return z;
      end; //


-- error1   FUNCTION myfun12 ended without RETURN 

Answer 4

我在下面遇到了同样的错误：

错误1320（42000）：在FUNCTION apple.addition中找不到RETURN

当我尝试创建

addition()

函数时，如下所示：

DELIMITER $$

CREATE FUNCTION addition(value INT) RETURNS INT 
DETERMINISTIC
BEGIN
UPDATE test SET num = num + value;
SELECT num INTO value FROM test;
END$$

DELIMITER ;

因此，我将 RETURN 语句设置为

addition()

函数，如下所示，然后我可以创建

addition()

函数而不会出现错误：

DELIMITER $$

CREATE FUNCTION addition(value INT) RETURNS INT 
DETERMINISTIC
BEGIN
UPDATE test SET num = num + value;
SELECT num INTO value FROM test;
RETURN value; -- Here
END$$

DELIMITER ;