日期:2014-05-18  浏览次数:20686 次

请问表的索引对应的列及列顺序存储在哪个系统表里?
或者怎样查询 谢谢

------解决方案--------------------
到sys.index_columns 和 sys.indexes 系统视图中去找找.
------解决方案--------------------
可這樣生成,這是之前寫的統計索引方法,你可查看ColName為你要的效果

SQL code
with Index1
as
(
select 
    top 100 percent row_number()over(partition by a.Name order by b.index_id) as ID,object_Name(a.object_id) as TableName,a.Name as IndexName,c.Name as ColName,
    description=a.type_desc,a.is_unique,a.is_primary_key,a.is_unique_constraint,OrderNr=CASE WHEN b.is_descending_key=0 THEN 'Asc' ELSE 'DESC' END


from 
    sys.indexes a 
join
    sys.index_columns b on a.Object_id=b.Object_id and a.index_id=b.index_id 
join
    sys.columns c on c.object_id=a.object_id and c.column_id=b.column_id
where 
    objectproperty(a.object_id,'IsUserTable')=1 and a.Object_id<>object_id('dtproperties')
order by TableName,a.Name
)
,index2
as
(
select 
    TableName,IndexName,ColName,is_unique,is_primary_key,is_unique_constraint,description
from 
    (select  distinct TableName,IndexName,is_unique,is_primary_key,is_unique_constraint,description from Index1)a
CROSS apply
    (select ColName=stuff((select ','+ColName +' '+OrderNr from Index1 where TableName=a.TableName and IndexName=a.IndexName order by ID for xml PATH('')),1,1,''))b
)
select 
    * 
from 
    index2
order by TableName,IndexName

------解决方案--------------------
SQL code
--结合sys.indexes和sys.index_columns,sys.objects,sys.columns查询索引所属的表或视图的信息
select
  o.name as 表名,
  i.name as 索引名,
  c.name as 列名,
  i.type_desc as 类型描述,
  is_primary_key as 主键约束,
  is_unique_constraint as 唯一约束,
  is_disabled as 禁用
from
  sys.objects o 
inner join
  sys.indexes i
on
  i.object_id=o.object_id
inner join 
  sys.index_columns ic
on
  ic.index_id=i.index_id and ic.object_id=i.object_id
inner join
  sys.columns c
on
  ic.column_id=c.column_id and ic.object_id=c.object_id
go



--查询索引的键和列信息
select 
  o.name as 表名,
  i.name as 索引名,
  c.name as 字段编号,
from
  sysindexes i inner join sysobjects o 
on
  i.id=o.id
inner join
  sysindexkeys k 
on
  o.id=k.id and i.indid=k.indid
inner join
  syscolumns c 
on
  c.id=i.id and k.colid=c.colid
where
  o.name='表名'