创建触发器帮助Oracle

Question

我正在尝试创建一个触发器，该触发器将执行以下操作。

在表A上插入后，根据ID（TableA.id = TableB.id）查询表B并将相应的信息插入到TableA中

我有种感觉，我还有很长的路要走，所以任何帮助都将不胜感激

CREATE OR REPLACE TRIGGER myTrig
AFTER INSERT
ON TABLEA
BEGIN
  INSERT INTO TABLEA
  SELECT TABLEB.FIRST_NAME, TABLEB.LAST_NAME, SYSDATE
  FROM TABLEA JOIN TABLEB ON 
  TABLEA.STUDENT_ID=TABLEB.STUDENT_ID
  insert into TABLEA values (....);
END;

Answer 1

这是行不通的。 只是想想你在问什么。 您需要一个将行插入tableA时触发的触发器，以将行插入tableA 。 扳机何时停止发射？

Oracle足够聪明，可以介入并防止触发器盘旋成无限大：

SQL> create or replace trigger t69_after_ins
  2      after insert on t69
  3  begin
  4      insert into t69 values ('blah', 'blah', 99);
  5  end;
  6  /

Trigger created.

SQL>

这是发生的事情：

SQL> insert into t69 values ('this', 'that', 1)      
   2  /

insert into t69 values ('this', 'that', 1)
            *
ERROR at line 1:
ORA-00036: maximum number of recursive SQL levels (50) exceeded
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_INS'
ORA-06512: at "APC.T69_AFTER_INS", line 2
ORA-04088: error during execution of trigger 'APC.T69_AFTER_I


SQL>

---

“给定两个表中的ID，是否有任何方法可以基于来自TableB的信息来更新TableA？”

如果“更新”实际上是指UPDATE，并且-关键-根据要实现的精确逻辑，则可能是：

SQL> create or replace trigger t69_after_ins
  2      after insert on t69
  3  begin
  4      update t69
  5          set name = ( select name from t23
  6                       where t23.id = t69.id )
  7          where name is null;
  8  end;
  9  /

Trigger created.

SQL> insert into t69 (id, name) values (122, null)
  2  /

1 row created.

SQL> select name from t69
  2  where id = 122
  3  /

NAME
----------
MAISIE

SQL>

但是，这仍然是一个坏主意。 触发器很难理解，并且可能对我们的SQL性能产生不利影响。 因此，我建议您尝试找出一种将逻辑构建到应用程序主体中的方法，而不要尝试使用触发器。