Question

我正在尝试创建一个触发器来执行以下操作。

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TableA

到目前为止，我有一种感觉，所以任何帮助都会受到赞赏

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

这不起作用。想想你在问什么。您想要一个触发器，当一行插入tableA到将一行插入tableA 时触发。触发器什么时候会停止发射？

Oracle非常聪明，可以介入并阻止触发器进入无限远：

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>

以下是发生的事情：

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>

“有什么办法可以更新 TableA基于TableB给出的信息两个表中都有一个id？ “

如果通过“更新”实际上意味着更新，并且 - 至关重要 - 根据您希望实现的精确逻辑，那么也许是：

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>

然而，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性能产生有害的影响。因此，我建议您尝试找出将逻辑构建到应用程序主体中的方法，而不是尝试使用触发器。

CREATE触发器帮助Oracle

1 个答案: