如何列出数据库中每个表的行数。相当于

select count(*) from table1
select count(*) from table2
...
select count(*) from tableN

我会张贴一个解决方案,但其他方法是受欢迎的


当前回答

select T.object_id, T.name, I.indid, I.rows 
  from Sys.tables T 
  left join Sys.sysindexes I 
    on (I.id = T.object_id and (indid =1 or indid =0 ))
 where T.type='U'

这里indid=1表示群集索引,而indid=0表示堆索引

其他回答

首先想到的是使用sp_msForEachTable

exec sp_msforeachtable 'select count(*) from ?'

但是它没有列出表名,所以可以扩展到

exec sp_msforeachtable 'select parsename(''?'', 1),  count(*) from ?'

这里的问题是,如果数据库有超过100个表,你会得到以下错误消息:

查询已超过最大值 可以的结果集的数目 显示在结果网格中。只有 前100个结果集是 显示在网格中。

所以我最终使用表变量来存储结果

declare @stats table (n sysname, c int)
insert into @stats
    exec sp_msforeachtable 'select parsename(''?'', 1),  count(*) from ?'
select 
    * 
from @stats
order by c desc

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。

        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

QUERY_RESULT_PHOTO

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: