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。
使用SQL Server 2016，这给出了所有索引的完整列表，并包含每个表的转储，以便您可以查看表之间的关系。它还显示包含在覆盖索引中的列:
select t.name TableName, i.name IdxName, c.name ColName
, ic.index_column_id ColPosition
, i.type_desc Type
, case when i.is_primary_key = 1 then 'Yes' else '' end [Primary?]
, case when i.is_unique = 1 then 'Yes' else '' end [Unique?]
, case when ic.is_included_column = 0 then '' else 'Yes - Included' end [CoveredColumn?]
, 'indexes >>>>' [*indexes*], i.*, 'index_columns >>>>' [*index_columns*]
, ic.*, 'tables >>>>' [*tables*]
, t.*, 'columns >>>>' [*columns*], c.*
from sys.index_columns ic
join sys.tables t on t.object_id = ic.object_id
join sys.columns c on c.object_id = t.object_id and c.column_id = ic.column_id
join sys.indexes i on i.object_id = t.object_id and i.index_id = ic.index_id
order by TableName, IdxName, ColPosition
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
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
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
