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。
当我有这个需求时，我使用了以下查询…
SELECT
TableName = t.name,
ColumnId = col.column_id,
ColumnName = col.name,
DataType = ty.name,
MaxSize = ty.max_length,
IsNullable = CASE WHEN (col.is_nullable = 1) THEN 'Y' END,
IsIdentity = CASE WHEN (col.is_identity = 1) THEN 'Y' END,
IsPrimaryKey = CASE WHEN (ic.column_id = col.column_id) THEN 'Y' END,
IsForeignKey = CASE WHEN (fkc.parent_column_id = col.column_id) THEN 'Y' END,
IsDefault = CASE WHEN (dc.parent_column_id = col.column_id) THEN 'Y' END
FROM
sys.tables t
INNER JOIN
sys.columns col ON t.object_id = col.object_id
LEFT JOIN
sys.indexes ind ON t.object_id = ind.object_id
LEFT JOIN
sys.index_columns ic ON ic.index_id=ind.index_id AND ic.object_id = col.object_id and ic.column_id = col.column_id
LEFT JOIN sys.foreign_key_columns fkc
ON fkc.parent_object_id = col.object_id AND fkc.parent_column_id=col.column_id
LEFT JOIN sys.default_constraints dc
ON dc.parent_object_id = col.object_id AND dc.parent_column_id=col.column_id
LEFT JOIN
sys.types ty on ty.user_type_id = col.user_type_id
WHERE
--t.name='<TABLENAME>'
t.schema_id = 10 --SCHEMA ID
AND ind.is_primary_key=1
ORDER BY
t.name, ColumnId
我想到了这个，它给了我我需要的准确的概述。有帮助的是，每个索引得到一行，索引列被聚合到其中。
select
o.name as ObjectName,
i.name as IndexName,
i.is_primary_key as [PrimaryKey],
SUBSTRING(i.[type_desc],0,6) as IndexType,
i.is_unique as [Unique],
Columns.[Normal] as IndexColumns,
Columns.[Included] as IncludedColumns
from sys.indexes i
join sys.objects o on i.object_id = o.object_id
cross apply
(
select
substring
(
(
select ', ' + co.[name]
from sys.index_columns ic
join sys.columns co on co.object_id = i.object_id and co.column_id = ic.column_id
where ic.object_id = i.object_id and ic.index_id = i.index_id and ic.is_included_column = 0
order by ic.key_ordinal
for xml path('')
)
, 3
, 10000
) as [Normal]
, substring
(
(
select ', ' + co.[name]
from sys.index_columns ic
join sys.columns co on co.object_id = i.object_id and co.column_id = ic.column_id
where ic.object_id = i.object_id and ic.index_id = i.index_id and ic.is_included_column = 1
order by ic.key_ordinal
for xml path('')
)
, 3
, 10000
) as [Included]
) Columns
where o.[type] = 'U' --USER_TABLE
order by o.[name], i.[name], i.is_primary_key desc
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
当我有这个需求时，我使用了以下查询…
SELECT
TableName = t.name,
ColumnId = col.column_id,
ColumnName = col.name,
DataType = ty.name,
MaxSize = ty.max_length,
IsNullable = CASE WHEN (col.is_nullable = 1) THEN 'Y' END,
IsIdentity = CASE WHEN (col.is_identity = 1) THEN 'Y' END,
IsPrimaryKey = CASE WHEN (ic.column_id = col.column_id) THEN 'Y' END,
IsForeignKey = CASE WHEN (fkc.parent_column_id = col.column_id) THEN 'Y' END,
IsDefault = CASE WHEN (dc.parent_column_id = col.column_id) THEN 'Y' END
FROM
sys.tables t
INNER JOIN
sys.columns col ON t.object_id = col.object_id
LEFT JOIN
sys.indexes ind ON t.object_id = ind.object_id
LEFT JOIN
sys.index_columns ic ON ic.index_id=ind.index_id AND ic.object_id = col.object_id and ic.column_id = col.column_id
LEFT JOIN sys.foreign_key_columns fkc
ON fkc.parent_object_id = col.object_id AND fkc.parent_column_id=col.column_id
LEFT JOIN sys.default_constraints dc
ON dc.parent_object_id = col.object_id AND dc.parent_column_id=col.column_id
LEFT JOIN
sys.types ty on ty.user_type_id = col.user_type_id
WHERE
--t.name='<TABLENAME>'
t.schema_id = 10 --SCHEMA ID
AND ind.is_primary_key=1
ORDER BY
t.name, ColumnId
我可以大胆回答这个饱和的问题吗?
这是@marc_s答案的自由重做，混合了来自@Tim Ford的一些东西，目标是有一个更干净和更简单的结果集和最终显示和排序，以满足我当前的需要。
SELECT
OBJECT_SCHEMA_NAME(t.[object_id],DB_ID()) AS [Schema],
t.[name] AS [TableName],
ind.[name] AS [IndexName],
col.[name] AS [ColumnName],
ic.column_id AS [ColumnId],
ind.[type_desc] AS [IndexTypeDesc],
col.is_identity AS [IsIdentity],
ind.[is_unique] AS [IsUnique],
ind.[is_primary_key] AS [IsPrimaryKey],
ic.[is_descending_key] AS [IsDescendingKey],
ic.[is_included_column] AS [IsIncludedColumn]
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
t.is_ms_shipped = 0
--ind.is_primary_key = 1 -- include or not pks, etc
--AND ind.is_unique = 0
--AND ind.is_unique_constraint = 0
ORDER BY
[Schema],
TableName,
IndexName,
[ColumnId],
ColumnName
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
