SELECT a.ROOM_NO,
TITLEDEED_BASIC.TITLEDEED_NO,
SUBSCRIBE_BASIC.SUBSCRIBE_NO
FROM TITLEDEED_BASIC
RIGHT OUTER JOIN ROOM_BASIC a ON TITLEDEED_BASIC.TITLEDEED_ID = a.TITLEDEED_ID, SUBSCRIBE_BASIC
RIGHT OUTER JOIN ROOM_BASIC b ON SUBSCRIBE_BASIC.SUBSCRIBE_ID = a.SUBSCRIBE_ID
TITLEDEED_BASIC.TITLEDEED_NO,
SUBSCRIBE_BASIC.SUBSCRIBE_NO
FROM TITLEDEED_BASIC
RIGHT OUTER JOIN ROOM_BASIC a ON TITLEDEED_BASIC.TITLEDEED_ID = a.TITLEDEED_ID, SUBSCRIBE_BASIC
RIGHT OUTER JOIN ROOM_BASIC b ON SUBSCRIBE_BASIC.SUBSCRIBE_ID = a.SUBSCRIBE_ID
SELECT a.ROOM_NO,
TITLEDEED_BASIC.TITLEDEED_NO,
SUBSCRIBE_BASIC.SUBSCRIBE_NO
FROM TITLEDEED_BASIC
RIGHT OUTER JOIN ROOM_BASIC AS a ON TITLEDEED_BASIC.TITLEDEED_ID = a.TITLEDEED_ID, SUBSCRIBE_BASIC
RIGHT OUTER JOIN ROOM_BASIC AS b ON SUBSCRIBE_BASIC.SUBSCRIBE_ID = a.SUBSCRIBE_ID出错如下(在SQL查询分析器中):
服务器: 消息 107,级别 16,状态 2,行 1
列前缀 'a' 与查询中所用的表名或别名不匹配。
TITLEDEED_BASIC.TITLEDEED_NO,
SUBSCRIBE_BASIC.SUBSCRIBE_NO
FROM TITLEDEED_BASIC
RIGHT OUTER JOIN ROOM_BASIC AS a ON TITLEDEED_BASIC.TITLEDEED_ID = a.TITLEDEED_ID, SUBSCRIBE_BASIC
RIGHT OUTER JOIN ROOM_BASIC AS b ON SUBSCRIBE_BASIC.SUBSCRIBE_ID = b.SUBSCRIBE_ID这个应该正确了。楼主试一下。你把‘RIGHT OUTER JOIN ROOM_BASIC AS b ON SUBSCRIBE_BASIC.SUBSCRIBE_ID = a.SUBSCRIBE_ID'中的'a.SUBSCRIBE_ID'改成‘b.SUBSCRIBE_ID'就行了