2

我有基于酒店房间住宿的 Oracle 数据库。我创建了一个程序 Room_taken，但我无法使其工作。

这是我的 SQL 过程代码：

CREATE
PROCEDURE Room_taken
(
v_guestID IN NUMBER DEFAULT NULL ,
v_stayfrom IN DATE DEFAULT NULL ,
v_stayto IN DATE DEFAULT NULL ,
v_roomID IN NUMBER DEFAULT NULL ,
v_paymentID IN NUMBER DEFAULT NULL ,
v_totalprice IN NUMBER DEFAULT NULL 
)
AS
v_nofrom DATE;
v_noto DATE;
v_idroom NUMBER(10,0);

BEGIN

BEGIN
SELECT Stay_from, Stay_to     
INTO v_nofrom, v_noto
FROM Stayover
WHERE ID_room = v_roomID;
IF NOT ( v_nofrom >= v_stayfrom
AND v_noto <= v_stayto ) THEN

BEGIN
INSERT INTO Stayover
( Stay_from, Stay_to, Total_price, ID_Room, ID_Guest, ID_Payment )
VALUES ( v_stayfrom, v_stayto, v_totalprice, v_roomID, v_guestID, v_paymentID );
END;
ELSE

BEGIN
raise_application_error( -20002, 'Room is taken!' );
END;
END IF;
END;
END;

当我输入以下参数后运行程序时：

DECLARE
v_guestID NUMBER;
v_stayfrom DATE;
v_stayto DATE;
v_roomID NUMBER;
v_paymentID NUMBER;
v_totalprice NUMBER;
BEGIN
v_guestID := 1;
v_stayfrom := '17.01.2012';
v_stayto := '19.01.2012';
v_roomID := 1;
v_paymentID := 1;
v_totalprice := 300;

Room_taken(
v_guestID => v_guestID,
v_stayfrom => v_stayfrom,
v_stayto => v_stayto,
v_roomid => v_roomid,
v_paymentID => v_paymentID,
v_totalprice => v_totalprice
);
END;

我收到一个错误：

Connecting to the database HOTEL.
ORA-01403: no data found
ORA-06512: at "HOTEL_ROOMS.ROOM_TAKEN", line 18
ORA-06512: at line 16
Process exited.
Disconnecting from the database HOTEL.

我究竟做错了什么？

4

2 回答 2

4

您收到错误是因为SELECT INTO构造必须在 PL/SQL 中准确返回 1 行。您的第一个选择不返回任何行，因此NO_DATA_FOUND会引发错误。

如果有 2 行（例如 1 月的预订，2 月的预订），TOO_MANY_ROWS则会引发错误。

相反，您要做的是确保插入时不会出现重叠。例如，以下查询将返回一段时间内的所有预订：

SELECT * 
  FROM Stayover 
 WHERE ID_room = v_roomID
   AND stay_from <= v_stayto 
   AND stay_to >= v_stayfrom

请注意，我比较stay_from了v_stayto，这不是错字。

您可以将上述查询包含在您的代码中，如下所示：

CREATE OR REPLACE PROCEDURE Room_taken(v_guestID    IN NUMBER,
                                       v_stayfrom   IN DATE,
                                       v_stayto     IN DATE,
                                       v_roomID     IN NUMBER,
                                       v_paymentID  IN NUMBER DEFAULT NULL,
                                       v_totalprice IN NUMBER DEFAULT NULL) AS
   l_nb_reservation NUMBER;
BEGIN
   SELECT COUNT(*)
     INTO l_nb_reservation
     FROM Stayover
    WHERE ID_room = v_roomID
      AND stay_from <= v_stayto
      AND stay_to >= v_stayfrom;
   IF l_nb_reservation > 0 THEN
      raise_application_error(-20002, 'Room is taken!');
   END IF;

   INSERT INTO Stayover
      (Stay_from, Stay_to, Total_price, ID_Room, ID_Guest, ID_Payment)
   VALUES
      (v_stayfrom, v_stayto, v_totalprice, v_roomID, v_guestID, v_paymentID);
END;

现在您每个时段只能预订一次房间：

SQL> EXEC Room_taken(1, trunc(SYSDATE), trunc(SYSDATE), 1);

PL/SQL procedure successfully completed

SQL> EXEC Room_taken(1, trunc(SYSDATE), trunc(SYSDATE), 1);

begin Room_taken(1, trunc(SYSDATE), trunc(SYSDATE), 1); end;

ORA-20002: Room is taken!
ORA-06512: à "APPS.ROOM_TAKEN", ligne 16
ORA-06512: à ligne 2
于 2013-02-26T16:42:41.300 回答
1

错误消息
ORA-01403：未找到数据

错误原因

您尝试了以下方法之一：

  1. 您执行了一个 SELECT INTO 语句，但没有返回任何行。
  2. 您引用了表中未初始化的行。
  3. 您使用 UTL_FILE 包读取了文件末尾。

显示建议的解决方案
检查此链接
http://www.techonthenet.com/oracle/errors/ora01403.php

于 2013-02-26T16:46:52.760 回答