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
以下是最好的方法:
SELECT sys.tables.object_id, sys.tables.name as table_name, sys.columns.name as column_name, sys.indexes.name as index_name,
sys.indexes.is_unique, sys.indexes.is_primary_key
FROM sys.tables, sys.indexes, sys.index_columns, sys.columns
WHERE (sys.tables.object_id = sys.indexes.object_id AND sys.tables.object_id = sys.index_columns.object_id AND sys.tables.object_id = sys.columns.object_id
AND sys.indexes.index_id = sys.index_columns.index_id AND sys.index_columns.column_id = sys.columns.column_id)
AND sys.tables.name = 'your_table_name'
我更喜欢使用隐式连接，因为它对我来说更容易理解。您可以删除object_id引用，因为您可能不需要它。
欢呼。
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
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
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:
