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

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

根据公认的答案和另外两个问题1,2，我整理了以下问题:

SELECT
    QUOTENAME(t.name) AS TableName,
    QUOTENAME(i.name) AS IndexName,
    i.is_primary_key,
    i.is_unique,
    i.is_unique_constraint,
    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, '') AS KeyColumns,
    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, '') AS IncludedColumns,
    u.user_seeks,
    u.user_scans,
    u.user_lookups,
    u.user_updates
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

该查询返回如下所示的结果，其中显示了索引的列表、它们的列和用法。非常有助于确定哪个指数比其他指数表现更好:

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

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