Question

我尝试按如下方式创建触发器：

CREATE OR REPLACE TRIGGER update_student
AFTER UPDATE OF school ON student FOR EACH ROW
BEGIN
    UPDATE school SET num2=num2+1 WHERE school.id=(SELECT school FROM inserted);
END;
/

但是当我执行代码时，它显示Trigger created with compilation errors，然后执行show errors;，它显示如下：

2/5 PL/SQL: SQL Statement ignored
2/71 PL/SQL: ORA-00942: table or view does not exist

我想这里的表格意味着学校，但我确信表school存在。问题出在哪里？请帮我一把。谢谢！

我修改了代码如下：

CREATE OR REPLACE TRIGGER update_student
AFTER UPDATE OF school ON student FOR EACH ROW
BEGIN
    UPDATE :new.school SET num2=num2+1 WHERE school.id=(SELECT :new.school FROM inserted);
END;
/

但是发生了新的错误。它显示：

2/5      PL/SQL: SQL Statement ignored
2/24     PL/SQL: ORA-00903: invalid table name

我该如何解决？谢谢！

Answer 1

在触发器中，您应该使用：NEW和：OLD伪行来访问新旧字段值。因此，您的UPDATE语句应为：

UPDATE school
  SET num2 = num2 + 1
  WHERE school.id = :NEW.SCHOOL;

您可能还想减少旧学校的学生人数：

UPDATE SCHOOL
  SET NUM2 = NUM2 - 1
  WHERE SCHOOL.ID = :OLD.SCHOOL;

如果STUDENT.SCHOOL（旧版本或新版本）为NULL，您应该避免尝试更新SCHOOL行：

CREATE OR REPLACE TRIGGER UPDATE_STUDENT
  AFTER UPDATE OF SCHOOL ON STUDENT
  FOR EACH ROW
BEGIN
  IF :NEW.SCHOOL IS NOT NULL THEN
    UPDATE SCHOOL
      SET NUM2 = NUM2 + 1
      WHERE SCHOOL.ID = :NEW.SCHOOL;
  END IF;

  IF :OLD.SCHOOL IS NOT NULL THEN
    UPDATE SCHOOL
      SET NUM2 = NUM2 - 1
      WHERE SCHOOL.ID = :OLD.SCHOOL;
  END IF;
END UPDATE_STUDENT;

祝你好运。

Oracle12c：使用编译错误创建的触发器（表或视图不存在）

1 个答案: