如何在SQL Server 2005+中获得所有索引和索引列的列表?我能想到的最接近的是:
select s.name, t.name, i.name, c.name from sys.tables t
inner join sys.schemas s on t.schema_id = s.schema_id
inner join sys.indexes i on i.object_id = t.object_id
inner join sys.index_columns ic on ic.object_id = t.object_id
inner join sys.columns c on c.object_id = t.object_id and
ic.column_id = c.column_id
where i.index_id > 0
and i.type in (1, 2) -- clustered & nonclustered only
and i.is_primary_key = 0 -- do not include PK indexes
and i.is_unique_constraint = 0 -- do not include UQ
and i.is_disabled = 0
and i.is_hypothetical = 0
and ic.key_ordinal > 0
order by ic.key_ordinal
这可不是我想要的。
我想要的是，列出所有用户定义的索引，(这意味着不支持唯一约束和主键的索引)与所有列(按它们在索引定义中的出现方式排序)以及尽可能多的元数据。
我没有经过，但是我在原作者发布的查询中得到了我想要的东西。
我使用它(没有条件/过滤器)来满足我的需求，但它给出了不正确的结果
主要问题是在index_id上没有连接条件的情况下得到叉乘
SELECT S.NAME SCHEMA_NAME,T.NAME TABLE_NAME,I.NAME INDEX_NAME,C.NAME COLUMN_NAME
FROM SYS.TABLES T
INNER JOIN SYS.SCHEMAS S
ON T.SCHEMA_ID = S.SCHEMA_ID
INNER JOIN SYS.INDEXES I
ON I.OBJECT_ID = T.OBJECT_ID
INNER JOIN SYS.INDEX_COLUMNS IC
ON IC.OBJECT_ID = T.OBJECT_ID
INNER JOIN SYS.COLUMNS C
ON C.OBJECT_ID = T.OBJECT_ID
**AND IC.INDEX_ID = I.INDEX_ID**
AND IC.COLUMN_ID = C.COLUMN_ID
WHERE 1=1
ORDER BY I.NAME,I.INDEX_ID,IC.KEY_ORDINAL
sELECT
TableName = t.name,
IndexName = ind.name,
--IndexId = ind.index_id,
ColumnId = ic.index_column_id,
ColumnName = col.name,
key_ordinal,
ind.type_desc
--ind.*,
--ic.*,
--col.*
FROM
sys.indexes ind
INNER JOIN
sys.index_columns ic ON ind.object_id = ic.object_id and ind.index_id = ic.index_id
INNER JOIN
sys.columns col ON ic.object_id = col.object_id and ic.column_id = col.column_id
INNER JOIN
sys.tables t ON ind.object_id = t.object_id
WHERE
ind.is_primary_key = 0
AND ind.is_unique = 0
AND ind.is_unique_constraint = 0
AND t.is_ms_shipped = 0
and t.name='CompanyReconciliation' --table name
and key_ordinal>0
ORDER BY
t.name, ind.name, ind.index_id, ic.index_column_id
根据Tim Ford的代码，这是正确答案:
select tab.[name] as [table_name],
idx.[name] as [index_name],
allc.[name] as [column_name],
idx.[type_desc],
idx.[is_unique],
idx.[data_space_id],
idx.[ignore_dup_key],
idx.[is_primary_key],
idx.[is_unique_constraint],
idx.[fill_factor],
idx.[is_padded],
idx.[is_disabled],
idx.[is_hypothetical],
idx.[allow_row_locks],
idx.[allow_page_locks],
idxc.[is_descending_key],
idxc.[is_included_column],
idxc.[index_column_id]
from sys.[tables] as tab
inner join sys.[indexes] idx on tab.[object_id] = idx.[object_id]
inner join sys.[index_columns] idxc on idx.[object_id] = idxc.[object_id] and idx.[index_id] = idxc.[index_id]
inner join sys.[all_columns] allc on tab.[object_id] = allc.[object_id] and idxc.[column_id] = allc.[column_id]
where tab.[name] Like '%table_name%'
and idx.[name] Like '%index_name%'
order by tab.[name], idx.[index_id], idxc.[index_column_id]
这是一种回退到索引的方法。您可以使用SHOWCONTIG来评估碎片。它将列出数据库或表的所有索引，以及统计信息。我要提醒的是，在大型数据库上，它可能是长时间运行的。对我来说，这种方法的好处之一是您不必是管理员就可以使用它。
——显示数据库中所有索引的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG WITH ALL_INDEXES
GO
.．.完成后关闭NOCOUNT
——显示表中所有索引的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors) WITH ALL_INDEXES
GO
——显示特定索引上的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors,aunmind)
GO
使用SQL Server 2016，这给出了所有索引的完整列表，并包含每个表的转储，以便您可以查看表之间的关系。它还显示包含在覆盖索引中的列:
select t.name TableName, i.name IdxName, c.name ColName
, ic.index_column_id ColPosition
, i.type_desc Type
, case when i.is_primary_key = 1 then 'Yes' else '' end [Primary?]
, case when i.is_unique = 1 then 'Yes' else '' end [Unique?]
, case when ic.is_included_column = 0 then '' else 'Yes - Included' end [CoveredColumn?]
, 'indexes >>>>' [*indexes*], i.*, 'index_columns >>>>' [*index_columns*]
, ic.*, 'tables >>>>' [*tables*]
, t.*, 'columns >>>>' [*columns*], c.*
from sys.index_columns ic
join sys.tables t on t.object_id = ic.object_id
join sys.columns c on c.object_id = t.object_id and c.column_id = ic.column_id
join sys.indexes i on i.object_id = t.object_id and i.index_id = ic.index_id
order by TableName, IdxName, ColPosition
with connect(schema_name,table_name,index_name,index_column_id,column_name) as
( select s.name schema_name, t.name table_name, i.name index_name, index_column_id, cast(c.name as varchar(max)) column_name
from sys.tables t
inner join sys.schemas s on t.schema_id = s.schema_id
inner join sys.indexes i on i.object_id = t.object_id
inner join sys.index_columns ic on ic.object_id = t.object_id and ic.index_id=i.index_id
inner join sys.columns c on c.object_id = t.object_id and
ic.column_id = c.column_id
where index_column_id=1
union all
select s.name schema_name, t.name table_name, i.name index_name, ic.index_column_id, cast(connect.column_name + ',' + c.name as varchar(max)) column_name
from sys.tables t
inner join sys.schemas s on t.schema_id = s.schema_id
inner join sys.indexes i on i.object_id = t.object_id
inner join sys.index_columns ic on ic.object_id = t.object_id and ic.index_id=i.index_id
inner join sys.columns c on c.object_id = t.object_id and
ic.column_id = c.column_id join connect on
connect.index_column_id+1 = ic.index_column_id
and connect.schema_name = s.name
and connect.table_name = t.name
and connect.index_name = i.name)
select connect.schema_name,connect.table_name,connect.index_name,connect.column_name
from connect join (select schema_name,table_name,index_name,MAX(index_column_id) index_column_id
from connect group by schema_name,table_name,index_name) mx
on connect.schema_name = mx.schema_name
and connect.table_name = mx.table_name
and connect.index_name = mx.index_name
and connect.index_column_id = mx.index_column_id
order by 1,2,3
