sql server递归子节点、父节点sql查询表结构的实例
2019/6/30 17:02:33
本文主要是介绍sql server递归子节点、父节点sql查询表结构的实例,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
一、查询当前部门下的所有子部门
WITH dept AS ( SELECT * FROM dbo.deptTab --部门表 WHERE pid = @id UNION ALL SELECT d.* FROM dbo.deptTab d INNER JOIN dept ON d.pid = dept.id ) SELECT * FROM dept
二、查询当前部门所有上级部门
WITH tab AS ( SELECT DepId , ParentId , DepName , [Enable] , 0 AS [Level] FROM deptTab WITH ( NOLOCK ) --表名 WHERE [Enable] = 1 AND depId = @depId UNION ALL SELECT b.DepId , b.ParentId , b.DepName , b.[Enable] , a.[Level] + 1 FROM tab a , deptTab b WITH ( NOLOCK ) WHERE a.ParentId = b.depId AND b.[enable] = 1 ) SELECT * FROM tab WITH ( NOLOCK ) WHERE [enable] = 1 ORDER BY [level] DESC
三、查询当前表的说明描述
SELECT tbs.name 表名 , ds.value 描述 FROM sys.extended_properties ds LEFT JOIN sysobjects tbs ON ds.major_id = tbs.id WHERE ds.minor_id = 0 AND tbs.name = 'userTab';--表名
四、查询当前表的表结构(字段名、属性、默认值、说明等)
SELECT CASE WHEN col.colorder = 1 THEN obj.name ELSE '' END AS 表名 , col.colorder AS 序号 , col.name AS 列名 , ISNULL(ep.[value], '') AS 列说明 , t.name AS 数据类型 , col.length AS 长度 , ISNULL(COLUMNPROPERTY(col.id, col.name, 'Scale'), 0) AS 小数位数 , CASE WHEN COLUMNPROPERTY(col.id, col.name, 'IsIdentity') = 1 THEN '√' ELSE '' END AS 标识 , CASE WHEN EXISTS ( SELECT 1 FROM dbo.sysindexes si INNER JOIN dbo.sysindexkeys sik ON si.id = sik.id AND si.indid = sik.indid INNER JOIN dbo.syscolumns sc ON sc.id = sik.id AND sc.colid = sik.colid INNER JOIN dbo.sysobjects so ON so.name = si.name AND so.xtype = 'PK' WHERE sc.id = col.id AND sc.colid = col.colid ) THEN '√' ELSE '' END AS 主键 , CASE WHEN col.isnullable = 1 THEN '√' ELSE '' END AS 允许空 , ISNULL(comm.text, '') AS 默认值 FROM dbo.syscolumns col LEFT JOIN dbo.systypes t ON col.xtype = t.xusertype INNER JOIN dbo.sysobjects obj ON col.id = obj.id AND obj.xtype = 'U' AND obj.status >= 0 LEFT JOIN dbo.syscomments comm ON col.cdefault = comm.id LEFT JOIN sys.extended_properties ep ON col.id = ep.major_id AND col.colid = ep.minor_id AND ep.name = 'MS_Description' LEFT JOIN sys.extended_properties epTwo ON obj.id = epTwo.major_id AND epTwo.minor_id = 0 AND epTwo.name = 'MS_Description' WHERE obj.name = 'userTab'--表名(点此修改) ORDER BY col.colorder;
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