如果您使用MySQL >4。X你可以用这个:

select TABLE_NAME, TABLE_ROWS from information_schema.TABLES where TABLE_SCHEMA="test";

请记住，对于某些存储引擎，TABLE_ROWS只是一个近似值。

你可以试试这个:

SELECT  OBJECT_SCHEMA_NAME(ps.object_Id) AS [schemaname],
        OBJECT_NAME(ps.object_id) AS [tablename],
        row_count AS [rows]
FROM sys.dm_db_partition_stats ps
WHERE OBJECT_SCHEMA_NAME(ps.object_Id) <> 'sys' AND ps.index_id < 2
ORDER BY 
        OBJECT_SCHEMA_NAME(ps.object_Id),
        OBJECT_NAME(ps.object_id)

SELECT 
    T.NAME AS 'TABLE NAME',
    P.[ROWS] AS 'NO OF ROWS'
FROM SYS.TABLES T 
INNER JOIN  SYS.PARTITIONS P ON T.OBJECT_ID=P.OBJECT_ID;

如果您使用MySQL >4。X你可以用这个:

select TABLE_NAME, TABLE_ROWS from information_schema.TABLES where TABLE_SCHEMA="test";

请记住，对于某些存储引擎，TABLE_ROWS只是一个近似值。

接受的答案在Azure SQL上不适合我，这里有一个，它非常快，完全符合我的要求:

select t.name, s.row_count
from sys.tables t
join sys.dm_db_partition_stats s
  ON t.object_id = s.object_id
    and t.type_desc = 'USER_TABLE'
    and t.name not like '%dss%'
    and s.index_id = 1
order by s.row_count desc

如果你使用的是SQL Server 2005及以上版本，你也可以使用这个:

SELECT 
    t.NAME AS TableName,
    i.name as indexName,
    p.[Rows],
    sum(a.total_pages) as TotalPages, 
    sum(a.used_pages) as UsedPages, 
    sum(a.data_pages) as DataPages,
    (sum(a.total_pages) * 8) / 1024 as TotalSpaceMB, 
    (sum(a.used_pages) * 8) / 1024 as UsedSpaceMB, 
    (sum(a.data_pages) * 8) / 1024 as DataSpaceMB
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
WHERE 
    t.NAME NOT LIKE 'dt%' AND
    i.OBJECT_ID > 255 AND   
    i.index_id <= 1
GROUP BY 
    t.NAME, i.object_id, i.index_id, i.name, p.[Rows]
ORDER BY 
    object_name(i.object_id) 

在我看来，它比sp_msforeachtable输出更容易处理。