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

这种方法使用字符串连接来动态生成包含所有表及其计数的语句，就像原始问题中给出的示例:

          SELECT COUNT(*) AS Count,'[dbo].[tbl1]' AS TableName FROM [dbo].[tbl1]
UNION ALL SELECT COUNT(*) AS Count,'[dbo].[tbl2]' AS TableName FROM [dbo].[tbl2]
UNION ALL SELECT...

最后用EXEC执行:

DECLARE @cmd VARCHAR(MAX)=STUFF(
                    (
                        SELECT 'UNION ALL SELECT COUNT(*) AS Count,''' 
                              + QUOTENAME(t.TABLE_SCHEMA) + '.' + QUOTENAME(t.TABLE_NAME) 
                              + ''' AS TableName FROM ' + QUOTENAME(t.TABLE_SCHEMA) + '.' + QUOTENAME(t.TABLE_NAME)
                        FROM INFORMATION_SCHEMA.TABLES AS t
                        WHERE TABLE_TYPE='BASE TABLE'
                        FOR XML PATH('')
                    ),1,10,'');
EXEC(@cmd);

sp_MSForEachTable 'DECLARE @t AS VARCHAR(MAX); 
SELECT @t = CAST(COUNT(1) as VARCHAR(MAX)) 
+ CHAR(9) + CHAR(9) + ''?'' FROM ? ; PRINT @t'

输出:

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;

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;