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。
首先，请注意，以上所有查询都可能遗漏或错误地合并索引的INCLUDE列。在某些情况下，还缺少列的正确排序和/或ASC/DESC选项。
由jona修改了上述查询。顺便说一句，在我使用的许多数据库中，我都安装了自己的CLR CONCATENATE聚合函数，因此下面的代码依赖于存在这样的东西。上面的SQL语句简化为更易于维护:
SELECT
s.[name] AS [schema_name]
, t.[name] AS [table_name]
, i.[name] AS [index_name]
, dbo.Concatenate(CASE WHEN ic.[key_ordinal] > 0 AND ic.[is_descending_key] = 1 THEN c.[name] + ' DESC' WHEN key_ordinal > 0 THEN c.[name] ELSE NULL END,',',1) AS [columns]
, dbo.Concatenate(CASE WHEN ic.[is_included_column] = 1 THEN c.[name] ELSE NULL END,',',1) AS [includes]
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
GROUP BY
s.[name]
, t.[name]
, i.[name]
ORDER BY
s.[name]
, t.[name]
, i.[name]
如果您的环境允许将基于clr的函数添加到其中，那么就会有许多级联聚合。
我需要得到特定的索引，它们的索引列和包含的列。以下是我使用的查询:
SELECT INX.[name] AS [Index Name]
,TBL.[name] AS [Table Name]
,DS1.[IndexColumnsNames]
,DS2.[IncludedColumnsNames]
FROM [sys].[indexes] INX
INNER JOIN [sys].[tables] TBL
ON INX.[object_id] = TBL.[object_id]
CROSS APPLY
(
SELECT STUFF
(
(
SELECT ' [' + CLS.[name] + ']'
FROM [sys].[index_columns] INXCLS
INNER JOIN [sys].[columns] CLS
ON INXCLS.[object_id] = CLS.[object_id]
AND INXCLS.[column_id] = CLS.[column_id]
WHERE INX.[object_id] = INXCLS.[object_id]
AND INX.[index_id] = INXCLS.[index_id]
AND INXCLS.[is_included_column] = 0
FOR XML PATH('')
)
,1
,1
,''
)
) DS1 ([IndexColumnsNames])
CROSS APPLY
(
SELECT STUFF
(
(
SELECT ' [' + CLS.[name] + ']'
FROM [sys].[index_columns] INXCLS
INNER JOIN [sys].[columns] CLS
ON INXCLS.[object_id] = CLS.[object_id]
AND INXCLS.[column_id] = CLS.[column_id]
WHERE INX.[object_id] = INXCLS.[object_id]
AND INX.[index_id] = INXCLS.[index_id]
AND INXCLS.[is_included_column] = 1
FOR XML PATH('')
)
,1
,1
,''
)
) DS2 ([IncludedColumnsNames])
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]
——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]
select i.object_id, i.name as [index] , STRING_AGG(c.name,', ') as [column], o.name as [table] from sys.indexes i
INNER join sys.index_columns ic on ic.object_id = i.object_id and ic.index_id = i.index_id
INNER join sys.columns c on c.object_id = ic.object_id and ic.column_id = c.column_id
INNER JOIN sys.objects o on o.object_id = i.object_id
where i.object_id > 100 and i.is_primary_key = 0 and i.is_unique = 0 and o.is_ms_shipped <> 1
group by i.object_id, i.name, o.name
order by i.name
将此用于sql 2016及更高级别，它会显示object_id, indexname，列和表名为非唯一的，没有主键
