Question

CREATE FUNCTION `new_function` ()
RETURNS bigint
BEGIN

set  @Ingredient:= (select Ingredient_cost_ingredientID  
from menu_has_ingredient_cost,bill_details_has_menu
where bill_details_has_menu.Menu_ItemID=menu_has_ingredient_cost.Menu_ItemID) 

RETURN Ingredient ;

Answer 1

你应该

RETURN @Ingredient ;

还需要一个

END;

其他/替代变更的

建议：

BEGIN
 DECLARE Ingredient BIGINT;

 select Ingredient_cost_ingredientID into Ingredient
 from 
    menu_has_ingredient_cost,
    bill_details_has_menu
 where 
    bill_details_has_menu.Menu_ItemID = 
     menu_has_ingredient_cost.Menu_ItemID;

 RETURN Ingredient;
END;

Answer 2

除了Ravinder的更正，我还会：

将WHERE中的加入替换为JOIN
抛弃局部变量 - 您可以直接返回结果
因为函数返回标量BIGINT，所以您需要确保查询永远不会返回多个值，例如：通过使用聚合。
您在RETURN

CREATE FUNCTION `new_function`()
RETURNS BIGINT
BEGIN
  RETURN
    (SELECT SUM(Ingredient_cost_ingredientID)  
      FROM menu_has_ingredient_cost 
        INNER JOIN bill_details_has_menu
        ON bill_details_has_menu.Menu_ItemID
          =menu_has_ingredient_cost.Menu_ItemID);
END;

Fiddle here

我无法在mysql中返回函数的值

2 个答案: