>>Check my blog post
>>http://blogs.lessthandot.com/index.php/DataMgmt/DataDesign/how-to-get-information-about-all-databas
>>
>>I believe I do have this script there (may be even a few).
>
>I executed the one "Record Count in every table in a database". It is very slow. I am better off just to scan all my tables one by one and execute the other script which provides instant result. Is that the script you were referring to?
>
>However, when I executed "Sizes of All Tables in a Database", that one gives an instant result and provides all the fields as the script I initially provided. That one seems to be the good one to use.

This record count should be quick:

DECLARE @SQL NVARCHAR(MAX)

 

SET @SQL = ''

--select * from sys.databases 

SELECT @SQL = @SQL + CHAR(13) + 'USE ' + QUOTENAME([name]) + ';

SELECT ' +QUOTENAME([name],'''') + 'as [Database Name], 

  SchemaName=s.name

  ,TableName=t.name

  ,CreateDate=t.create_date

  ,ModifyDate=t.modify_date

  ,p.rows

  ,DataInKB=sum(a.used_pages)*8

FROM sys.schemas s

JOIN sys.tables t on s.schema_id=t.schema_id

JOIN sys.partitions p on t.object_id=p.object_id

JOIN sys.allocation_units a on a.container_id=p.partition_id

GROUP BY s.name, t.name, t.create_date, t.modify_date, p.rows

ORDER BY SchemaName, TableName' FROM sys.databases  

 

EXECUTE (@SQL)