Question

我正在尝试编写一个触发器，将表中的所有条目从1到10排列（最大值具有等级10，最小值具有等级1，所有其他值都被赋予其间的整数值）。这是触发器代码：

DELIMITER $$
CREATE TRIGGER risks_before_insert
BEFORE INSERT ON risks
FOR EACH ROW
BEGIN

DECLARE max_cost double; #previous maximum expected_cost in project
DECLARE min_cost double; #previous minimum expected_cost in project
DECLARE slope double; #slope for prioritizing functioN

SELECT MAX(expected_cost), MIN(expected_cost) INTO max_cost, min_cost FROM view_risks; #GET EXTREME VALUES FROM TABLE, STORE IN MEMORY

/*
Update Priority Monetary Rankings
*/
IF (max_cost IS NULL OR min_cost IS NULL) THEN #check for empty table
    SET NEW.priority_monetary = 10;
ELSE IF ((NEW.expected_cost <= max_cost) AND (NEW.expected_cost >= min_cost)) THEN #NEW VALUE DOES NOT CHANGE TABLE EXTREMES
    IF (max_cost – min_cost = 0) THEN
        SET NEW.priority_monetary = 10;
    ELSE
        SET slope = 9 / (max_cost – min_cost);
        SET NEW.priority_monetary = slope * (NEW.expected_cost - min_cost) + 1;
    END IF;
ELSE IF (NEW.expected_cost > max_cost) THEN
    SET NEW.priority_monetary = 10;
    SET slope = 9 / (NEW.expected_cost – min_cost);
    UPDATE TABLE risks SET priority_monetary = slope * (expected_cost - min_cost) + 1;
ELSE #NEW VALUE CORRESPONDS TO A MINIMUM
    SET NEW.priority_monetary = 1;
    SET slope = 9 / (max_cost – NEW.expected_cost);
    UPDATE TABLE risks SET priority_monetary = slope * (expected_cost – NEW.min_cost) + 1;
END IF;
END $$
DELIMITER ;

但是，我收到以下错误：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '– min_cost); SET NEW.priority_monetary = slope * (NEW.expected_cost - min_cost' at line 22

任何人都可以解释什么是错的吗？供您参考，我的排名算法由以下函数描述，常量为maxval和minval：

Rank（x）= 1 + slope（x - minval），其中slope = 9 /（maxval - minval）。

Answer 1

你应该检查一个超出范围的例外。它永远不应该等于0.如果它确实会查询失败并填满数据库。

在将maxval设置为minval <之前，您应该检查maxval - minval和0以确保slope不等于9/(maxval - minval) / p>

如果你运行这个

SET slope = 9 / (max_cost – min_cost);

并尝试将slope设置为9/0它会崩溃。

Answer 2

我不熟悉触发器，但我认为您的问题出在SET声明中。

我在想你不能设置NEW.priority_monetary

我之前没有看到任何类似SQL的语法，但我仍然是游戏的新手，但我猜你如果放弃NEW.而只是设置priority_monetary那么错误可能会消失

这个MySQL触发器有什么问题？

2 个答案: