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。
SQL Server 2014工作解决方案。我在这里只包含了少量的输出字段,但您可以随意添加任何您喜欢的字段。
SELECT
o.object_id AS objectId
,o.name AS objectName
,i.index_id AS indexId
,i.name AS indexName
,i.type_desc AS typeDesc
,ic.index_column_id AS indexColumnId
,ic.key_ordinal AS keyOrdinal
,ic.is_included_column AS isIncludedColumn
,ic.column_id AS columnId
,c.name AS columnName
FROM {database}.sys.objects AS o
INNER JOIN {database}.sys.columns AS c ON
c.object_id = o.object_id
AND o.type = 'U'
INNER JOIN {database}.sys.indexes AS i ON
i.object_id = o.object_id
INNER JOIN {database}.sys.index_columns AS ic ON
ic.object_id = i.object_id
AND ic.index_id = i.index_id
AND ic.column_id = c.column_id
ORDER BY
o.object_id
,i.index_id
,ic.index_column_id
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
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
正确的一个在这里(当我们在一个表上有多个索引时,以上所有帖子都会给出笛卡尔积结果)
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
AND i.index_id = ic.index_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
AND t.name = 'DimCustomer'
order by ic.key_ordinal