Home > Database > Mysql Tutorial > Oracle|sql server|access 数据库里的所有表名,字段名

Oracle|sql server|access 数据库里的所有表名,字段名

WBOY
Release: 2016-06-07 17:48:13
Original
1203 people have browsed it

Oracle|sql server|access 里的所有表名,字段名


* from user_tables where table_name = '用户名'

如果是用该用户登录使用以下语句:
SELECT *
FROM USER_TABLES;
如果是用其他用户:
SELECT * FROM ALL_TABLES WHERE OWNER='USER_NAME' 
首先,第一句:是查询的该用户下的所有表吗?一般的普通用户,只是授予了connect 和 resource权限,也可以查看部分系统表吗?
怎样查看该用户【以该用户登录】下的自己创建的所有表呢?
其次,第二句,当我以sys身份登录时,怎么不可以使用这一句呢?SELECT *
FROM ALL_TABLES WHERE OWNER='xiaoming'【xiaoming是我自己创建的用户,只是授予了connect 和 resource权限】

SQL SERVER
查看所有表名:
select   name   from   sysobjects   where   type='U'

查询表的所有字段名:
Select name from syscolumns Where ID=OBJECT_ID('表名')

select * from information_schema.tables
select * from information_schema.views
select * from information_schema.columns

 

ACCESS

查看所有表名:
select   name   from   MSysObjects   where   type=1   and   flags=0

 MSysObjects是系统对象,默认情况是隐藏的。通过工具、选项、视图、显示、系统对象可以使之显示出来。 


参考:sql server获取库名,表名


1.获取表的基本字段属性

--获取SqlServer中表结构
SELECT syscolumns.name,systypes.name,syscolumns.isnullable,
syscolumns.length
FROM syscolumns, systypes
WHERE syscolumns.xusertype = systypes.xusertype
AND syscolumns.id = object_id('你的表名')

2.获取字段的描述信息

--获取SqlServer中表结构 主键,及描述
declare @table_name as varchar(max)
set @table_name = '你的表名'
select sys.columns.name, sys.types.name, sys.columns.max_length, sys.columns.is_nullable,
  (select count(*) from sys.identity_columns where sys.identity_columns.object_id = sys.columns.object_id and sys.columns.column_id = sys.identity_columns.column_id) as is_identity ,
  (select value from sys.extended_properties where sys.extended_properties.major_id = sys.columns.object_id and sys.extended_properties.minor_id = sys.columns.column_id) as description
  from sys.columns, sys.tables, sys.types where sys.columns.object_id = sys.tables.object_id and sys.columns.system_type_id=sys.types.system_type_id and order by sys.columns.column_id

3.单独查询表的递增字段

--单独查询表递增字段
select [name] from syscolumns where
id=object_id(N'你的表名') and COLUMNPROPERTY(id,name,'IsIdentity')=1

4.获取表的主外键

--获取表主外键约束
exec sp_helpconstraint   '你的表名' ;

5.相当完整的表结构查询

--很全面的表结构
exec sp_helpconstraint   '你的表名' ;

SELECT 表名  = CASE a.colorder WHEN 1 THEN c.name ELSE '' END,

       序    = a.colorder,

       字段名= a.name,

       标识  = CASE COLUMNPROPERTY(a.id,a.name,'IsIdentity') WHEN 1 THEN '√' ELSE '' END,

       主键  = CASE

  WHEN EXISTS ( SELECT * FROM sysobjects WHERE xtype='PK'

AND name IN (SELECT [name] FROM sysindexes   WHERE id=a.id

  AND indid IN (SELECT indid FROM sysindexkeys WHERE id=a.id

 AND colid IN (SELECT colid FROM syscolumns   WHERE id=a.id

  AND name=a.name)))) THEN '√' ELSE '' END,

     类型= b.name,

     字节数= a.length,

     长度  = COLUMNPROPERTY(a.id,a.name,'Precision'),

     小数  = CASE ISNULL(COLUMNPROPERTY(a.id,a.name,'Scale'),0) WHEN 0 THEN '' ELSE CAST(COLUMNPROPERTY(a.id,a.name,'Scale') AS VARCHAR) END,

     允许空= CASE a.isnullable WHEN 1 THEN '√' ELSE '' END,

     默认值= ISNULL(d.[text],''),

     说明  = ISNULL(e.[value],'')

 FROM syscolumns a

 LEFT JOIN systypes b ON a.xtype=b.xusertype

INNER JOIN sysobjects c ON a.id=c.id AND c.xtype='U' AND c.name'dtproperties'

LEFT  JOIN syscomments d ON a.cdefault=d.id

LEFT  JOIN sys.extended_properties e ON a.id=e.class AND a.colid=e.minor_id

ORDER BY c.name, a.colorder

6.获取所有的库名

--获取服务器中的所遇库名

select * from master..sysdatabases

7.获取服务器上所有库的所有表

--获取服务器上所有库的所有表名

use master

declare @db_name varchar(100)

declare @sql varchar(200)

declare cur_tables cursor

for

select name from sysdatabases /*where name like 'by_%'*/

 

open cur_tables

fetch next from cur_tables into @db_name

while @@fetch_status = 0

begin

--set @db_name = @db_name + '.dbo.sysobjects'

print @db_name

set @sql = 'select * from ' + @db_name + '.dbo.sysobjects where xtype =''U'''

exec (@sql)

fetch next from cur_tables into @db_name

end

close cur_tables

deallocate cur_tables

go

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template