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。
对于每个索引的唯一列:
select s.name, t.name, i.name, i.index_id,c.name,c.column_id
from sys.schemas s
inner join sys.tables t 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 i.object_id = object_id('previous.account_1')
order by index_id,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
根据公认的答案和另外两个问题1,2，我整理了以下问题:
SELECT
QUOTENAME(t.name) AS TableName,
QUOTENAME(i.name) AS IndexName,
i.is_primary_key,
i.is_unique,
i.is_unique_constraint,
STUFF(REPLACE(REPLACE((
SELECT QUOTENAME(c.name) + CASE WHEN ic.is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
FROM sys.index_columns AS ic
INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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
), '<row>', ', '), '</row>', ''), 1, 2, '') AS KeyColumns,
STUFF(REPLACE(REPLACE((
SELECT QUOTENAME(c.name) AS [data()]
FROM sys.index_columns AS ic
INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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.index_column_id
FOR XML PATH
), '<row>', ', '), '</row>', ''), 1, 2, '') AS IncludedColumns,
u.user_seeks,
u.user_scans,
u.user_lookups,
u.user_updates
FROM sys.tables AS t
INNER JOIN sys.indexes AS i ON t.object_id = i.object_id
LEFT JOIN sys.dm_db_index_usage_stats AS u ON i.object_id = u.object_id AND i.index_id = u.index_id
WHERE t.is_ms_shipped = 0
AND i.type <> 0
该查询返回如下所示的结果，其中显示了索引的列表、它们的列和用法。非常有助于确定哪个指数比其他指数表现更好:
——简短而甜蜜:
SELECT OBJECT_SCHEMA_NAME(T.[object_id],DB_ID()) AS [Schema],
T.[name] AS [table_name], I.[name] AS [index_name], AC.[name] AS [column_name],
I.[type_desc], I.[is_unique], I.[data_space_id], I.[ignore_dup_key], I.[is_primary_key],
I.[is_unique_constraint], I.[fill_factor], I.[is_padded], I.[is_disabled], I.[is_hypothetical],
I.[allow_row_locks], I.[allow_page_locks], IC.[is_descending_key], IC.[is_included_column]
FROM sys.[tables] AS T
INNER JOIN sys.[indexes] I ON T.[object_id] = I.[object_id]
INNER JOIN sys.[index_columns] IC ON I.[object_id] = IC.[object_id]
INNER JOIN sys.[all_columns] AC ON T.[object_id] = AC.[object_id] AND IC.[column_id] = AC.[column_id]
WHERE T.[is_ms_shipped] = 0 AND I.[type_desc] <> 'HEAP'
ORDER BY T.[name], I.[index_id], IC.[key_ordinal]
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]
对于每个索引的唯一列:
select s.name, t.name, i.name, i.index_id,c.name,c.column_id
from sys.schemas s
inner join sys.tables t 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 i.object_id = object_id('previous.account_1')
order by index_id,column_id
