mysql - 无法创建返回某些值的存储例程

Question

我在 Maria DB 工作。我有存储过程。现在我需要添加一些返回功能。为此，我创建了一个新函数，如下所示 -

CREATE FUNCTION `insertChild`(`nodeId` INT, `newNodeName` INT)
    RETURNS INT
    LANGUAGE SQL
    NOT DETERMINISTIC
    CONTAINS SQL
    SQL SECURITY DEFINER
    COMMENT ''
BEGIN

    Rollback;

START TRANSACTION;

    SELECT @myRight := rgt FROM item_tree WHERE item_id = nodeId;

    UPDATE item_tree SET rgt = rgt + 2 WHERE rgt >= @myRight;
    UPDATE item_tree SET lft = lft + 2 WHERE lft > @myRight;
    INSERT INTO item_tree(item_name, lft, rgt) VALUES(newNodeName, @myRight , @myRight + 1);

    RETURN (SELECT (COUNT(parent.item_name) - 1) AS depth FROM item_tree AS node,
        item_tree AS parent
    WHERE node.item_id = last_insert_id() AND 
    node.lft BETWEEN parent.lft AND parent.rgt 
    GROUP BY node.item_name
    ORDER BY node.lft);

COMMIT;

END

但是，这是行不通的。它正在返回 SQL 错误 (1415):不允许从函数返回结果集。

这里有什么问题？

提前致谢!

Answer 1

我改变了它返回的方式。在变量中设置计数并返回它。请检查下面更新的代码

CREATE FUNCTION `insertChild`(`nodeId` INT, `newNodeName` INT)
    RETURNS INT
    LANGUAGE SQL
    NOT DETERMINISTIC
    CONTAINS SQL
    SQL SECURITY DEFINER
    COMMENT ''
BEGIN

    Rollback;

START TRANSACTION;

    SELECT @myRight := rgt FROM item_tree WHERE item_id = nodeId;

    UPDATE item_tree SET rgt = rgt + 2 WHERE rgt >= @myRight;
    UPDATE item_tree SET lft = lft + 2 WHERE lft > @myRight;
    INSERT INTO item_tree(item_name, lft, rgt) VALUES(newNodeName, @myRight , @myRight + 1);

    SELECT @myCount := (COUNT(parent.item_name) - 1) AS depth FROM item_tree AS node,
        item_tree AS parent
    WHERE node.item_id = last_insert_id() AND 
    node.lft BETWEEN parent.lft AND parent.rgt 
    GROUP BY node.item_name
    ORDER BY node.lft;

    RETURN @myCount;

COMMIT;

END