WITH Tree AS ( SELECT * FROM TableT AS tt WHERE tt.ID=8 UNION ALL SELECT tt.* FROM Tree AS t INNER JOIN TableT AS tt ON t.ID = tt.pCatID ) SELECT * FROM Tree AS t
查出父ID为8的所有节点下的数据
数据库树形结构查询语句
原文：http://www.cnblogs.com/hantianwei/p/4137023.html
