ORACLE SQL觸發器 - 未找到數據

Question

我創建了一個觸發器，以便在插入或更新預訂記錄時，如果評估爲0，則將預訂的詳細信息插入到審計表中。下面的觸發器完美的作品，如果我更新記錄，但是在插入它給了我下面的錯誤：

INSERT INTO "SCOTT"."BOOKING" (PASSENGER_PASSENGER_ID, VOYAGE_VOYAGE_ID, CABIN_NUM, CLASS, EVALUATION) VALUES ('2', '1', '202', 'SECOND', '0') 
ORA-01403: no data found 
ORA-01403: no data found 
ORA-06512: at "SCOTT.EVALUATION_TRIG", line 8 
ORA-04088: error during execution of trigger 'SCOTT.EVALUATION_TRIG' 
ORA-06512: at line 1 


One error saving changes to table "SCOTT"."BOOKING": 
Row 2: ORA-01403: no data found 
ORA-01403: no data found 
ORA-06512: at "SCOTT.EVALUATION_TRIG", line 8 
ORA-04088: error during execution of trigger 'SCOTT.EVALUATION_TRIG' 
ORA-06512: at line 1 

Oracle的SQL觸發代碼：

create or replace trigger EVALUATION_TRIG 
AFTER INSERT OR UPDATE 
ON BOOKING 
FOR EACH ROW 
WHEN (NEW.EVALUATION =0) 
DECLARE 
PRAGMA AUTONOMOUS_TRANSACTION; 
PASSENGER_NAME VARCHAR2(30); 
CRUISE_NAME VARCHAR2(30); 
VOYAGE_DATE DATE; 
SHIP_NAME1 VARCHAR2(30); 
BEGIN 
SELECT NAME INTO PASSENGER_NAME FROM PASSENGER JOIN BOOKING 
ON PASSENGER.PASSENGER_ID = BOOKING.PASSENGER_PASSENGER_ID 
WHERE BOOKING.PASSENGER_PASSENGER_ID = :NEW.PASSENGER_PASSENGER_ID 
AND ROWNUM = 1; 
SELECT NAME INTO CRUISE_NAME FROM CRUISE JOIN VOYAGE 
ON CRUISE.CRUISE_ID = VOYAGE.CRUISE_CRUISE_ID 
JOIN BOOKING ON 
VOYAGE.VOYAGE_ID = BOOKING.VOYAGE_VOYAGE_ID 
WHERE BOOKING.VOYAGE_VOYAGE_ID = :NEW.VOYAGE_VOYAGE_ID 
AND ROWNUM= 1; 
SELECT START_DATE INTO VOYAGE_DATE FROM VOYAGE JOIN BOOKING 
ON VOYAGE.VOYAGE_ID= BOOKING.VOYAGE_VOYAGE_ID 
WHERE BOOKING.VOYAGE_VOYAGE_ID = :NEW.VOYAGE_VOYAGE_ID 
AND ROWNUM = 1; 
SELECT SHIP_NAME INTO SHIP_NAME1 FROM SHIP JOIN VOYAGE 
ON SHIP.SHIP_ID = VOYAGE.SHIP_SHIP_ID JOIN BOOKING 
ON VOYAGE.VOYAGE_ID= BOOKING.VOYAGE_VOYAGE_ID 
WHERE BOOKING.VOYAGE_VOYAGE_ID = :NEW.VOYAGE_VOYAGE_ID 
AND ROWNUM = 1; 
INSERT INTO EVALUATION_AUDIT VALUES (PASSENGER_NAME, CRUISE_NAME,VOYAGE_DATE, SHIP_NAME1,:NEW.EVALUATION); 
COMMIT; 
END; 

Answer 1

當你執行一個select..into聲明出現該錯誤不會返回一行，例如觸發器中的第一個。

SELECT NAME INTO PASSENGER_NAME FROM PASSENGER JOIN BOOKING 
ON PASSENGER.PASSENGER_ID = BOOKING.PASSENGER_PASSENGER_ID 
WHERE BOOKING.PASSENGER_PASSENGER_ID = :NEW.PASSENGER_PASSENGER_ID 
AND ROWNUM = 1; 

這些新的值應填寫在刀片一樣好更新，但這並不意味着實際的數據同樣存在。

上面的查詢將在預訂時加入乘客，但是如果您爲該乘客插入第一個預訂，則不存在當前預訂。此外，要獲取乘客的姓名，根本不需要預訂。而且，在同一個表的行級觸發器上查詢表X可能會產生問題。

所以，長話短說，我覺得上面這句話應該是這樣的：

SELECT NAME INTO PASSENGER_NAME 
FROM PASSENGER 
WHERE PASSENGER.PASSENGER_ID = :NEW.PASSENGER_PASSENGER_ID; 

同樣的變化還需要其他的語句來進行。

之後，理論上你仍然會得到相同的錯誤，但只有在以下情況下：NEW.PASSENGER_PASSENGER_ID與任何乘客的ID不匹配。但是在那種情況下，我認爲如果你配置了這些約束，你應該首先得到一個外鍵約束錯誤。

我也看到在觸發器的末尾提交。這也沒有多大意義。觸發器是聲明的一部分，聲明本身應該是原子的。從觸發器提交，如果這可以工作，則會自動提交多行插入中的每一行。 Suggested reading material。