MS Access:WHERE-EXISTS子句不能处理视图?

时间:2012-01-31 13:07:47

标签: sql database ms-access select sql-view

先决条件:在MS Access 2010中,创建以下表格:

CREATE TABLE ATBL(ID INT);
INSERT INTO ATBL(ID) VALUES (1);
INSERT INTO ATBL(ID) VALUES (2);
INSERT INTO ATBL(ID) VALUES (3);

CREATE TABLE BTBL(ID INT);
INSERT INTO BTBL(ID) VALUES (1);
INSERT INTO BTBL(ID) VALUES (2);

还创建一个名为BVIEW的视图,它使用以下SELECT语句:

SELECT A.ID FROM ATBL AS A WHERE A.ID = 1 OR A.ID = 2

现在BVIEW应该具有与BTBL相同的内容。然而,以下两个查询将返回不同的结果:

SELECT A.ID FROM ATBL AS A WHERE EXISTS (SELECT 1 FROM  BTBL AS B WHERE B.ID=A.ID)
SELECT A.ID FROM ATBL AS A WHERE EXISTS (SELECT 1 FROM BVIEW AS B WHERE B.ID=A.ID)

第一个查询返回两个记录(1和2),但第二个查询返回ATBL的所有记录。这有什么不对?我错过了什么吗?

1 个答案:

答案 0 :(得分:6)

视图实际上是一个保存的SQL SELECT语句。至少,这是MS Access中保存的视图。你使用相同的内部变量A和B.恕我直言,他们正在混合。最后一行看起来像是

SELECT A.ID FROM ATBL AS A WHERE EXISTS (SELECT 1 FROM (SELECT A.ID FROM ATBL AS A WHERE A.ID = 1 OR A.ID = 2) AS B WHERE B.ID=A.ID)

尝试更改一些内部名称,例如:

SELECT AA.ID FROM ATBL AS AA WHERE AA.ID = 1 OR AA.ID = 2

所以,最后一行看起来像

SELECT A.ID FROM ATBL AS A WHERE EXISTS (SELECT 1 FROM (SELECT AA.ID FROM ATBL AS AA WHERE AA.ID = 1 OR AA.ID = 2) AS B WHERE B.ID=A.ID)

因此,正如我们在这里看到的,MS Access甚至不知道如何隔离别名!