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。
with connect(schema_name,table_name,index_name,index_column_id,column_name) as
( select s.name schema_name, t.name table_name, i.name index_name, index_column_id, cast(c.name as varchar(max)) 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 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 index_column_id=1
union all
select s.name schema_name, t.name table_name, i.name index_name, ic.index_column_id, cast(connect.column_name + ',' + c.name as varchar(max)) 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 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 join connect on
connect.index_column_id+1 = ic.index_column_id
and connect.schema_name = s.name
and connect.table_name = t.name
and connect.index_name = i.name)
select connect.schema_name,connect.table_name,connect.index_name,connect.column_name
from connect join (select schema_name,table_name,index_name,MAX(index_column_id) index_column_id
from connect group by schema_name,table_name,index_name) mx
on connect.schema_name = mx.schema_name
and connect.table_name = mx.table_name
and connect.index_name = mx.index_name
and connect.index_column_id = mx.index_column_id
order by 1,2,3
sELECT
TableName = t.name,
IndexName = ind.name,
--IndexId = ind.index_id,
ColumnId = ic.index_column_id,
ColumnName = col.name,
key_ordinal,
ind.type_desc
--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
and t.name='CompanyReconciliation' --table name
and key_ordinal>0
ORDER BY
t.name, ind.name, ind.index_id, ic.index_column_id
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
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
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
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