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。


当前回答

这是我的,工作在一个默认模式,但它可以很容易地改进 它提供了3列SQLQueries -创建/删除/重建(没有重组)

查询:

SELECT
'CREATE ' + 
CASE WHEN is_primary_key=1 THEN 'CLUSTERED' 
WHEN is_primary_key=0 and is_unique_constraint=0 THEN 'NONCLUSTERED'
WHEN is_primary_key=0 and is_unique_constraint=1 THEN 'UNIQUE' END  
+ ' INDEX ' +
QUOTENAME(i.name) + ' ON ' +
QUOTENAME(t.name) + ' ( '  + 
STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) + CASE WHEN ic.is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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
    ), '<row>', ', '), '</row>', ''), 1, 2, '') + ' ) '  -- keycols
+ COALESCE(' INCLUDE ( ' +
    STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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.index_column_id
        FOR XML PATH
    ), '<row>', ', '), '</row>', ''), 1, 2, '') + ' ) ',    -- included cols
    '') as [Create],
'DROP INDEX ' + QUOTENAME(i.name) + ' ON ' + QUOTENAME(t.name) as [Drop],
'ALTER INDEX ' + QUOTENAME(i.name)  + ' ON ' +QUOTENAME(t.name) + ' REBUILD ' as [Rebuild]
FROM sys.tables AS t
INNER JOIN sys.indexes AS i ON t.object_id = i.object_id
LEFT JOIN sys.dm_db_index_usage_stats AS u ON i.object_id = u.object_id AND i.index_id = u.index_id
WHERE t.is_ms_shipped = 0
AND i.type <> 0
order by QUOTENAME(t.name), is_primary_key desc

输出

Create                                                                                                      Drop                                    Rebuild
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CREATE CLUSTERED INDEX [PK_Table1] ON [Table1] ( [Tab1_ID] )                                                DROP INDEX [PK_Table1] ON [Table1]      ALTER INDEX [PK_Table1] ON [Table1] REBUILD 
CREATE UNIQUE INDEX [IX_Table1_Name] ON [Table1] ( [Tab1_Name] )                                            DROP INDEX [IX_Table1_Name] ON [Table1] ALTER INDEX [IX_Table1_Name] ON [Table1] REBUILD 
CREATE NONCLUSTERED INDEX [IX_Table2] ON [Table2] ( [Tab2_Name], [Tab2_City] )  INCLUDE ( [Tab2_PhoneNo] )  DROP INDEX [IX_Table2] ON [Table2]      ALTER INDEX [IX_Table2] ON [Table2] REBUILD

其他回答

——简短而甜蜜:

SELECT OBJECT_SCHEMA_NAME(T.[object_id],DB_ID()) AS [Schema],  
  T.[name] AS [table_name], I.[name] AS [index_name], AC.[name] AS [column_name],  
  I.[type_desc], I.[is_unique], I.[data_space_id], I.[ignore_dup_key], I.[is_primary_key], 
  I.[is_unique_constraint], I.[fill_factor],    I.[is_padded], I.[is_disabled], I.[is_hypothetical], 
  I.[allow_row_locks], I.[allow_page_locks], IC.[is_descending_key], IC.[is_included_column] 
FROM sys.[tables] AS T  
  INNER JOIN sys.[indexes] I ON T.[object_id] = I.[object_id]  
  INNER JOIN sys.[index_columns] IC ON I.[object_id] = IC.[object_id] 
  INNER JOIN sys.[all_columns] AC ON T.[object_id] = AC.[object_id] AND IC.[column_id] = AC.[column_id] 
WHERE T.[is_ms_shipped] = 0 AND I.[type_desc] <> 'HEAP' 
ORDER BY T.[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]
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

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;

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