2013-04-05 61 views
2

我正在尝试用光标编写mySQL过程来计算票价。我通过stationid,然后我找出他们在哪个区域。票价是1美元的设定值,并且每个区域的额外费用为0.20美元。 我到目前为止的代码已经运行,但是游标存在问题将值提取到变量中。mySQL存储过程 - 光标错误

任何帮助将不胜感激。由于

表:

DROP DATABASE IF EXISTS luasSystem; 
    CREATE DATABASE luasSystem; 
    USE luasSystem; 

    CREATE TABLE IF NOT EXISTS line 
    (
    line_id INT NOT NULL AUTO_INCREMENT, 
    Line_colour CHAR(10) NOT NULL, 
    PRIMARY KEY (line_id) 
    ) ENGINE=InnoDB; 

    CREATE TABLE IF NOT EXISTS zone 
    (
    zone_id INT NOT NULL AUTO_INCREMENT, 
    zone_name VARCHAR(20) NOT NULL, 
     line INT NOT NULL, 
    PRIMARY KEY (zone_id), 
    FOREIGN KEY (line) REFERENCES line(line_id) ON UPDATE CASCADE ON DELETE RESTRICT 
    ) ENGINE=InnoDB; 

    CREATE TABLE IF NOT EXISTS station 
    (
    station_id INT NOT NULL AUTO_INCREMENT, 
    station_name CHAR(20) NOT NULL, 
    service CHAR(20), 
     line INT NOT NULL, 
     zone INT NOT NULL, 
    PRIMARY KEY (station_id), 
    FOREIGN KEY (line) REFERENCES line(line_id) ON UPDATE CASCADE ON DELETE RESTRICT, 
    FOREIGN KEY (zone) REFERENCES zone(zone_id) ON UPDATE CASCADE ON DELETE RESTRICT 
    ) ENGINE=InnoDB; 

存储过程:你使用十进制值,而不是整数

DROP PROCEDURE IF EXISTS calculateFare; 
    DELIMITER // 
    CREATE PROCEDURE calculateFare 
    (
    IN stationid1 INT, IN stationid2 INT 
    ) 
    BEGIN 

    DECLARE zoneNum1 INT; 
    DECLARE zoneNum2 INT; 
    DECLARE num INT; 
    DECLARE fare DOUBLE; 

    DECLARE tableEnd BOOLEAN; 

    DECLARE zoneCur CURSOR FOR 
    SELECT zone, zone FROM station 
    WHERE station_name = stationid1 AND station_name = stationid2; 

    DECLARE CONTINUE HANDLER FOR NOT FOUND 
    SET tableEnd = TRUE; 

    OPEN zoneCur; 
    the_loop: LOOP 

    FETCH zoneCur 
    INTO zoneNum1, zoneNum2; 

    IF tableEnd THEN 
    CLOSE zoneCur; 
    LEAVE the_loop; 
    END IF; 

    SET fare = 1; 
    SET num = 0; 

    IF zoneNum1 < zoneNum2 THEN 
    SET num = zoneNum2 - zoneNum1; 
    ELSEIF zoneNum1 > zoneNum2 THEN 
    SET num = zoneNum1 - zoneNum2; 
    END IF; 

    SET fare = (num * 0.20) + 1; 
    SELECT fare; 
    END LOOP the_loop; 
    END // 
    DELIMITER ; 

    CAll calculateFare(3,5); 
+0

这看起来很奇怪(你的意思是把它们命名为相同?):SELECT zone,zone FROM station WHERE station_name = stationid1 AND station_name = stationid2; – ethrbunny 2013-04-05 20:51:19

+0

嗨,感谢您的回复。我试图让这两个电台“stationid1”和“stationid2”的区域,它不会允许我做多个选择语句。 – d199224 2013-04-05 20:55:50

+0

除非两个'stationid'值相等,否则这将永远不会返回任何内容。你正在和这两个条款。 – ethrbunny 2013-04-05 20:58:24

回答