作为marc_s答案（已被接受的答案）的一个简单扩展，它被调整为返回列计数并允许过滤：

SELECT *
FROM
(

SELECT 
    t.NAME AS TableName,
    s.Name AS SchemaName,
    p.rows AS RowCounts,
    COUNT(DISTINCT c.COLUMN_NAME) as ColumnCount,
    SUM(a.total_pages) * 8 AS TotalSpaceKB, 
    (SUM(a.used_pages) * 8) AS UsedSpaceKB, 
    (SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB
FROM 
    sys.tables t
INNER JOIN      
    sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN 
    sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
INNER JOIN 
    sys.allocation_units a ON p.partition_id = a.container_id
INNER JOIN
    INFORMATION_SCHEMA.COLUMNS c ON t.NAME = c.TABLE_NAME
LEFT OUTER JOIN 
    sys.schemas s ON t.schema_id = s.schema_id
WHERE 
    t.NAME NOT LIKE 'dt%' 
    AND t.is_ms_shipped = 0
    AND i.OBJECT_ID > 255
GROUP BY 
    t.Name, s.Name, p.Rows
) AS Result

WHERE
    RowCounts > 1000
    AND ColumnCount > 10
ORDER BY 
    UsedSpaceKB DESC

我发现这个查询很容易使用和快速。

select schema_name(tab.schema_id) + '.' + tab.name as [table], 
cast(sum(spc.used_pages * 8)/1024.00 as numeric(36, 2)) as used_mb,
cast(sum(spc.total_pages * 8)/1024.00 as numeric(36, 2)) as allocated_mb
from sys.tables (nolock) tab
inner join sys.indexes (nolock) ind 
    on tab.object_id = ind.object_id
inner join sys.partitions  (nolock) part 
    on ind.object_id = part.object_id and ind.index_id = part.index_id
inner join sys.allocation_units (nolock) spc
    on part.partition_id = spc.container_id
group by schema_name(tab.schema_id) + '.' + tab.name
order by sum(spc.used_pages) desc

SELECT 
    t.NAME AS TableName,
    s.Name AS SchemaName,
    p.rows,
    SUM(a.total_pages) * 8 AS TotalSpaceKB, 
    CAST(ROUND(((SUM(a.total_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS TotalSpaceMB,
    SUM(a.used_pages) * 8 AS UsedSpaceKB, 
    CAST(ROUND(((SUM(a.used_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS UsedSpaceMB, 
    (SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB,
    CAST(ROUND(((SUM(a.total_pages) - SUM(a.used_pages)) * 8) / 1024.00, 2) AS NUMERIC(36, 2)) AS UnusedSpaceMB
FROM 
    sys.tables t
INNER JOIN      
    sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN 
    sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
INNER JOIN 
    sys.allocation_units a ON p.partition_id = a.container_id
LEFT OUTER JOIN 
    sys.schemas s ON t.schema_id = s.schema_id
WHERE 
    t.NAME NOT LIKE 'dt%' 
    AND t.is_ms_shipped = 0
    AND i.OBJECT_ID > 255 
GROUP BY 
    t.Name, s.Name, p.Rows
ORDER BY 
    TotalSpaceMB DESC, t.Name

对于Azure，我使用了：

您应该有SSMS v17.x+

我使用；

正如用户Sparrow所提到的：

打开数据库>并选择表，然后按F7键你应该看到行数作为：

此处的SSMS连接到Azure数据库

如果需要精确计算SSMS中“表财产-存储”页面上的相同数字，则需要使用与SSMS中相同的方法进行计数（适用于sql server 2005及更高版本……也适用于具有LOB字段的表，因为仅计算“used_pages”不足以显示准确的索引大小）：

;with cte as (
SELECT
t.name as TableName,
SUM (s.used_page_count) as used_pages_count,
SUM (CASE
            WHEN (i.index_id < 2) THEN (in_row_data_page_count + lob_used_page_count + row_overflow_used_page_count)
            ELSE lob_used_page_count + row_overflow_used_page_count
        END) as pages
FROM sys.dm_db_partition_stats  AS s 
JOIN sys.tables AS t ON s.object_id = t.object_id
JOIN sys.indexes AS i ON i.[object_id] = t.[object_id] AND s.index_id = i.index_id
GROUP BY t.name
)
select
    cte.TableName, 
    cast((cte.pages * 8.)/1024 as decimal(10,3)) as TableSizeInMB, 
    cast(((CASE WHEN cte.used_pages_count > cte.pages 
                THEN cte.used_pages_count - cte.pages
                ELSE 0 
          END) * 8./1024) as decimal(10,3)) as IndexSizeInMB
from cte
order by 2 desc

从使用OSQL的命令提示符：

OSQL -E -d <*databasename*> -Q "exec sp_msforeachtable 'sp_spaceused [?]'" > result.txt