¿Hay alguna forma de averiguar el tamaño/tamaño de las tablas de la base de datos?

Question

tengo alrededor de 10 tablas en mi servidor sql 2008.

Actualmente, mi mdf es de alrededor de 3.5Gig. (También tengo algunos datos binarios en algunas de las tablas). Entonces, me pregunto si hay alguna manera de ver qué tablas son las más grandes.

¿Esto es posible?

¿Tal vez es un índice o un catálogo de FTS en su lugar?

Answer 1

plazo esto:

/****************************************************************************** 
** File: “GetTableSpaceUseage.sql” 
** Name: Get Table Space Useage for a specific schema 
** Auth: Robert C. Cain 
** Date: 01/27/2008 
** 
** Desc: Calls the sp_spaceused proc for each table in a schema and returns 
**  the Table Name, Number of Rows, and space used for each table. 
** 
** Called by: 
**  n/a – As needed 
** 
** Input Parameters: 
**  In the code check the value of @schemaname, if you need it for a 
**  schema other than dbo be sure to change it. 
** 
** Output Parameters: 
**  NA 
*******************************************************************************/ 

/*—————————————————————————*/ 
/* Drop the temp table if it's there from a previous run      */ 
/*—————————————————————————*/ 
if object_id(N'tempdb..[#TableSizes]') is not null 
    drop table #TableSizes ; 
go 

/*—————————————————————————*/ 
/* Create the temp table              */ 
/*—————————————————————————*/ 
create table #TableSizes 
    (
    [Table Name] nvarchar(128) /* Name of the table */ 
    , [Number of Rows] char(11) /* Number of rows existing in the table. */ 
    , [Reserved Space] varchar(18) /* Reserved space for table. */ 
    , [Data Space] varchar(18) /* Amount of space used by data in table. */ 
    , [Index Size] varchar(18) /* Amount of space used by indexes in table. */ 
    , [Unused Space] varchar(18) /* Amount of space reserved but not used. */ 
) ; 
go 

/*—————————————————————————*/ 
/* Load the temp table              */ 
/*—————————————————————————*/ 
declare @schemaname varchar(256) ; 
-- Make sure to set next line to the Schema name you want! 
set @schemaname = 'dbo' ; 

-- Create a cursor to cycle through the names of each table in the schema 
declare curSchemaTable cursor 
    for select sys.schemas.name + '.' + sys.objects.name 
     from sys.objects 
      , sys.schemas 
     where object_id > 100 
       and sys.schemas.name = @schemaname 
       /* For a specific table uncomment next line and supply name */ 
       --and sys.objects.name = 'specific-table-name-here'  
       and type_desc = 'USER_TABLE' 
       and sys.objects.schema_id = sys.schemas.schema_id ; 

open curSchemaTable ; 
declare @name varchar(256) ; /* This holds the name of the current table*/ 

-- Now loop thru the cursor, calling the sp_spaceused for each table 
fetch curSchemaTable into @name ; 
while (@@FETCH_STATUS = 0) 
    begin  
    insert into #TableSizes 
      exec sp_spaceused @objname = @name ;  
    fetch curSchemaTable into @name ; 
    end 

/* Important to both close and deallocate! */ 
close curSchemaTable ;  
deallocate curSchemaTable ; 


/*—————————————————————————*/ 
/* Feed the results back              */ 
/*—————————————————————————*/ 
select [Table Name] 
     , [Number of Rows] 
     , [Reserved Space] 
     , [Data Space] 
     , [Index Size] 
     , [Unused Space] 
from [#TableSizes] 
order by [Table Name] ; 

/*—————————————————————————*/ 
/* Remove the temp table              */ 
/*—————————————————————————*/ 
drop table #TableSizes ; 

taken de Robert Caine el blog

Editado el código para analizar, varios caracteres que estaban en comilla simple utiliza una comilla simple especial, además de la -- firmar.

Este código es para Microsoft SQL 2005 +

Answer 2

exec sp_spaceused <tablename> 

Answer 3

exec sp_spaceused [tablename] 

Answer 4

veces corro esto ... Se pone todas las tablas de tabla temporal, a través de los bucles y obtiene tamaños para todas las tablas. Los datos del resultado están en @ tablesize, por lo que puede consultarlo como lo desee.

Obras en SQL Server> 2005

declare @tables TABLE 
(
    table_name nvarchar(200) 
) 

declare @tablesizes TABLE 
(
    [name] nvarchar(200), 
    [rows] int, 
    reserved nvarchar(200), 
    data nvarchar(200), 
    index_size nvarchar(200), 
    unused nvarchar(200), 
    reserved_int int, 
    data_int int, 
    index_size_int int, 
    unused_int int 
) 

declare @t nvarchar(200) 

insert into @tables 
select Table_Name from information_schema.tables 

while exists(select * from @tables) 
begin 
    set @t=(select top 1 table_name from @tables) 

    insert into @tablesizes([name],[rows],reserved,data,index_size,unused) 
    exec sp_spaceused @t 

    delete top (1) from @tables 
end 

update @tablesizes set 
reserved_int=convert(int, replace(reserved,' KB','')), 
data_int=convert(int, replace(data,' KB','')), 
index_size_int=convert(int, replace(index_size,' KB','')), 
unused_int=convert(int, replace(unused,' KB','')) 

select * from @tablesizes order by data_int desc 

Answer 5

sys.allocations_units tiene la información que necesita. Se une con sys.partitions para agrupar todas las unidades de asignación de una partición y también para obtener el object_id más útil en lugar del esotérico allocation_unit_id.

select object_name(p.object_id), 
    sum(au.total_pages)*8 as [space_in_kb] 
    from sys.partitions p 
    join sys.allocation_units au on p.hobt_id = au.container_id 
    group by p.object_id 
    order by [space_in_kb] desc 

Y sí, todas las tablas (heap o agrupadas) son 'partitions', los términos no se refiere a tablas con particiones. sys.partitions también tiene la columna 'filas' que puede interesarte.

Answer 6

Esta consulta muestra el tamaño de cada tabla en la base de datos actual.

SELECT sysobjects.[name] AS [TableName], 
    SUM(sysindexes.reserved) * 8 AS [Size(KB)], 
    SUM(sysindexes.dpages) * 8 AS [Data(KB)], 
    (SUM(sysindexes.used) - SUM(sysindexes.dpages)) * 8 AS [Indexes(KB)], 
    (SUM(sysindexes.reserved) - SUM(sysindexes.dpages)) * 8 AS [Unused(KB)] 
FROM dbo.sysindexes AS sysindexes 
    JOIN dbo.sysobjects AS sysobjects ON sysobjects.id = sysindexes.id 
WHERE sysobjects.[type] = 'U' 
GROUP BY sysobjects.[name] 
ORDER BY [Size(KB)] DESC 

Answer 7

En SQL 2008+: haga clic en el nombre de base de datos en SSMS, seleccione Reports, entonces Standard Reports, entonces Disk Usage by Table.

Answer 8

que puede usar:

SELECT @@servername; 

IF EXISTS(SELECT name FROM tempdb.sys.tables WHERE name LIKE '#spaceUsed%') 
    BEGIN 
     DROP TABLE #spaceUsed; 
    END; 
CREATE TABLE #spaceUsed (
name VARCHAR(255) , 
rows INT , 
reserved VARCHAR(50) , 
data VARCHAR(50) , 
index_size VARCHAR(50) , 
unused VARCHAR(50)); 

EXEC sp_msforeachtable 
@command1 =' 
-- 
INSERT INTO #spaceUsed 
exec sp_spaceused N''?''; 
' 
,@whereand = ' And Object_id In (Select Object_id From sys.objects 
Where SCHEMA_NAME(Schema_ID) like ''%'')'; 

DECLARE 
    @spaceUsedData TABLE (
    name VARCHAR(255) , 
    rows INT , 
    reservedMB BIGINT NULL , 
    dataMB BIGINT NULL , 
    index_sizeMB BIGINT NULL , 
    unusedMB BIGINT NULL); 

INSERT INTO INTO @spaceUsedData (name , rows , reservedMB , dataMB ,index_sizeMB ,unusedMB) 
SELECT     name , rows , 
Convert (BIGINT ,Ltrim(Rtrim(Replace(reserved ,'KB' ,''))))/1024 , 
Convert (BIGINT ,Ltrim(Rtrim(Replace(data ,'KB' ,''))))/1024 , 
Convert (BIGINT ,Ltrim(Rtrim(Replace(index_size ,'KB' ,''))))/1024 , 
Convert (BIGINT ,Ltrim(Rtrim(Replace(unused ,'KB' ,''))))/1024 
FROM #spaceUsed; 

SELECT * , reservedMB+ dataMB+index_sizeMB+unusedMB AS TotalMB FROM @spaceUsedData 
ORDER BY rows DESC;