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。
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。
有两个“sys”目录视图可以参考:Indexes和sys.index_columns。
这些会给你关于下标和它们的列的任何信息。
编辑:这个查询非常接近你所寻找的:
SELECT
TableName = t.name,
IndexName = ind.name,
IndexId = ind.index_id,
ColumnId = ic.index_column_id,
ColumnName = col.name,
ind.*,
ic.*,
col.*
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
ind.is_primary_key = 0
AND ind.is_unique = 0
AND ind.is_unique_constraint = 0
AND t.is_ms_shipped = 0
ORDER BY
t.name, ind.name, ind.index_id, ic.is_included_column, ic.key_ordinal;
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
当我有这个需求时，我使用了以下查询…
SELECT
TableName = t.name,
ColumnId = col.column_id,
ColumnName = col.name,
DataType = ty.name,
MaxSize = ty.max_length,
IsNullable = CASE WHEN (col.is_nullable = 1) THEN 'Y' END,
IsIdentity = CASE WHEN (col.is_identity = 1) THEN 'Y' END,
IsPrimaryKey = CASE WHEN (ic.column_id = col.column_id) THEN 'Y' END,
IsForeignKey = CASE WHEN (fkc.parent_column_id = col.column_id) THEN 'Y' END,
IsDefault = CASE WHEN (dc.parent_column_id = col.column_id) THEN 'Y' END
FROM
sys.tables t
INNER JOIN
sys.columns col ON t.object_id = col.object_id
LEFT JOIN
sys.indexes ind ON t.object_id = ind.object_id
LEFT JOIN
sys.index_columns ic ON ic.index_id=ind.index_id AND ic.object_id = col.object_id and ic.column_id = col.column_id
LEFT JOIN sys.foreign_key_columns fkc
ON fkc.parent_object_id = col.object_id AND fkc.parent_column_id=col.column_id
LEFT JOIN sys.default_constraints dc
ON dc.parent_object_id = col.object_id AND dc.parent_column_id=col.column_id
LEFT JOIN
sys.types ty on ty.user_type_id = col.user_type_id
WHERE
--t.name='<TABLENAME>'
t.schema_id = 10 --SCHEMA ID
AND ind.is_primary_key=1
ORDER BY
t.name, ColumnId
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。
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]
