要在SQL Management Studio中获取该信息，右键单击数据库，然后选择Reports—> Standard Reports—> Disk Usage by Table。

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

这个sql脚本给出了所选数据库中每个表的模式、表名和行数:

SELECT SCHEMA_NAME(schema_id) AS [SchemaName],
[Tables].name AS [TableName],
SUM([Partitions].[rows]) AS [TotalRowCount]
FROM sys.tables AS [Tables]
JOIN sys.partitions AS [Partitions]
ON [Tables].[object_id] = [Partitions].[object_id]
AND [Partitions].index_id IN ( 0, 1 )
-- WHERE [Tables].name = N'name of the table'
GROUP BY SCHEMA_NAME(schema_id), [Tables].name
order by [TotalRowCount] desc

裁判:https://blog.sqlauthority.com/2017/05/24/sql-server-find-row-count-every-table-database-efficiently/

另一种表达方式:

SELECT  o.NAME TABLENAME,
  i.rowcnt 
FROM sysindexes AS i
  INNER JOIN sysobjects AS o ON i.id = o.id 
WHERE i.indid < 2  AND OBJECTPROPERTY(o.id, 'IsMSShipped') = 0
ORDER BY i.rowcnt desc

以下是我对这个问题的看法。它包含所有模式，只列出带行的表。YMMV

select distinct schema_name(t.schema_id) as schema_name, t.name as 
table_name, p.[Rows]
from sys.tables as t
INNER JOIN sys.indexes as 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
where p.[Rows] > 0
order by schema_name;

查找SQL reference (http://www.codeproject.com/Tips/811017/Fastest-way-to-find-row-count-of-all-tables-in-SQL)中所有表的行数的最快方法

SELECT T.name AS [TABLE NAME], I.rows AS [ROWCOUNT] 
    FROM   sys.tables AS T 
       INNER JOIN sys.sysindexes AS I ON T.object_id = I.id 
       AND I.indid < 2 
ORDER  BY I.rows DESC

        SELECT ( Schema_name(A.schema_id) + '.' + A.NAME ) AS TableName,
 Sum(B.rows)AS RecordCount 
    FROM   sys.objects A INNER JOIN sys.partitions B 
    ON A.object_id = B.object_id WHERE  A.type = 'U' 
        GROUP  BY A.schema_id,A.NAME ;

QUERY_PHOTO

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