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。
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]
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
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
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]
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
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
