這個MySQL觸發器排名有什麼問題？

Question

我想寫一個觸發器，將表中所有條目從1到10進行排序（最大值爲10，最小值爲1，所有其他值都在其間分配整數值）。下面是觸發代碼：

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

DECLARE n_project_id integer; #project_id associated with risk 
DECLARE max_cost double; #previous maximum expected_cost in project 
DECLARE min_cost double; #previous minimum expected_cost in project 
DECLARE max_impact double; #previous maximum impact_effect in project 
DECLARE min_impact double; #previous minimum impact_effect in project 
DECLARE slope double; #slope for prioritizing function 

SELECT t.project_id INTO n_project_id FROM tasks t WHERE t.task_id = NEW.task_id; #GET PROJECT_ID ASSOCIATED WITH THE RISK 

SET NEW.expected_cost = NEW.probability * NEW.cost_impact, NEW.overall_impact = NEW.probability * NEW.impact_effect; #CALCULATE EXPECTED_COST AND OVERALL_IMPACT FIELDS 

SELECT MAX(expected_cost), MIN(expected_cost), MAX(overall_impact), MIN(overall_impact) INTO max_cost, min_cost, max_impact, min_impact FROM view_risks WHERE r.project_id = n_project_id; #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; 
ELSEIF ((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; 
ELSEIF (NEW.expected_cost > max_cost) THEN 
    SET NEW.priority_monetary = 10; 
    SET slope = 9/(NEW.expected_cost - min_cost); 
    UPDATE risks SET priority_monetary = slope * (expected_cost - min_cost) + 1 WHERE project_id = n_project_id; 
ELSE #NEW VALUE CORRESPONDS TO A MINIMUM 
    SET NEW.priority_monetary = 1; 
    SET slope = 9/(max_cost - NEW.expected_cost); 
    UPDATE risks SET priority_monetary = slope * (expected_cost - min_cost) + 1 WHERE project_id = n_project_id; 
END IF; 

/* 
Update Priority Effect Rankings 
*/ 
IF (max_impact IS NULL OR min_impact IS NULL) THEN #check for empty table 
    SET NEW.priority_effect = 10; 
ELSEIF ((NEW.overall_impact <= max_impact) AND (NEW.overall_impact >= min_impact)) THEN 
#NEW VALUE DOES NOT CHANGE TABLE EXTREMES 
    IF (max_cost - min_cost = 0) THEN 
     SET NEW.priority_effect = 10; 
    ELSE 
     SET slope = 9/(max_impact - min_impact); 
SET NEW.priority_effect = slope * (NEW.overall_impact - min_impact) + 1; 
    END IF; 
ELSEIF (NEW.overall_impact > max_impact) THEN 
    SET NEW.priority_effect = 10; 
    SET slope = 9/(NEW.overall_impact - min_impact); 
    UPDATE risks SET priority_effect = slope * (overall_impact - min_impact) + 1 WHERE project_id = n_project_id; 
ELSE #NEW VALUE CORRESPONDS TO A MINIMUM 
    SET NEW.priority_effect = 1; 
    SET slope = 9/(max_impact - NEW.overall_impact); 
    UPDATE risks SET priority_effect = slope * (overall_impact - min_impact) + 1 WHERE project_id = n_project_id; 
END IF; 
END 
DELIMITER ; 

不過，我收到以下錯誤：

#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 '' at line 5 

誰能解釋一下什麼是錯的？爲了便於參考，我們的排名算法由以下函數用常數maxval和minval來描述：

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

謝謝！

附加信息：

字段類型： expected_cost =>十進制（11,2） priority_monetary => TINYINT（2）

示例值： expected_cost => 1000.00 priority_monetary => 2

Answer 1

你有ELSE IF你應該在哪裏使用ELSEIF。

同樣@Mihai寫道：你的缺點不會出現作爲缺點，並且你有UPDATE後的關鍵字TABLE（應該只有UPDATE tableName）。

PS我建議你使用MySql的一些工具，如MySql Workbench，它有Windows，大多數linux系統和OSX的發行版。它會讓你工作更好，並檢查你的語法。

更改後的代碼：

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; 
ELSEIF ((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; 
ELSEIF (NEW.expected_cost > max_cost) THEN 
    SET NEW.priority_monetary = 10; 
    SET slope = 9/(NEW.expected_cost - min_cost); 
    UPDATE 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 risks SET priority_monetary = slope * (expected_cost - NEW.min_cost) + 1; 
END IF; 

END $$ 
DELIMITER ; 

Answer 2

在我的MySql客戶端中，您的缺點並不會顯示爲不足，也做了一些小修改（UPDATE tableName without the keyword TABLE）

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_costs - 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 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 risks SET priority_monetary = slope * (expected_cost - NEW.min_cost) + 1; 
END IF; 
END $$ 
DELIMITER ;