sql server - vista - ¿Cómo se determina qué tablas SQL tienen una columna de identidad mediante programación?

modificar columna auto_increment sql server

Question

Quiero crear una lista de columnas en SQL Server 2005 que tengan columnas de identidad y su tabla correspondiente en T-SQL.

Los resultados serían algo así como:

TableName, ColumnName

Answer 1

Creo que esto funciona para SQL 2000:

SELECT
CASE WHEN C.autoval IS NOT NULL THEN
''Identity''
ELSE
''Not Identity''
AND
FROM
sysobjects O
INNER JOIN
syscolumns C
ON
O.id = C.id
WHERE
O.NAME = @TableName
AND
C.NAME = @ColumnName

Answer 2

En SQL 2005:

select object_name(object_id), name
from sys.columns
where is_identity = 1

Answer 3

Esta consulta parece hacer el truco:

SELECT
sys.objects.name AS table_name,
sys.columns.name AS column_name
FROM sys.columns JOIN sys.objects
ON sys.columns.object_id=sys.objects.object_id
WHERE
sys.columns.is_identity=1
AND
sys.objects.type in (N''U'')

Answer 4

Esto funcionó para SQL Server 2005, 2008 y 2012. Encontré que sys.identity_columns no contenía todas mis tablas con columnas de identidad.

SELECT a.name AS TableName, b.name AS IdentityColumn
FROM sys.sysobjects a
JOIN sys.syscolumns b
ON a.id = b.id
WHERE is_identity = 1
ORDER BY name;

En cuanto a la página de documentación, también se puede utilizar la columna de estado. También puede agregar el identificador de cuatro partes y funcionará en diferentes servidores.

SELECT a.name AS TableName, b.name AS IdentityColumn
FROM [YOUR_SERVER_NAME].[YOUR_DB_NAME].sys.sysobjects a
JOIN [YOUR_SERVER_NAME].[YOUR_DB_NAME].sys.syscolumns b
ON a.id = b.id
WHERE is_identity = 1
ORDER BY name;

Fuente: https://msdn.microsoft.com/en-us/library/ms186816.aspx

Answer 5

Esto funcionó para mí usando Sql Server 2008:

USE <database_name>;
GO
SELECT SCHEMA_NAME(schema_id) AS schema_name
, t.name AS table_name
, c.name AS column_name
FROM sys.tables AS t
JOIN sys.identity_columns c ON t.object_id = c.object_id
ORDER BY schema_name, table_name;
GO

Answer 6

Lista de tablas sin columna de identidad basada en la respuesta de Guillermo :

SELECT DISTINCT TABLE_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE (TABLE_SCHEMA = ''dbo'') AND (OBJECTPROPERTY(OBJECT_ID(TABLE_NAME), ''TableHasIdentity'') = 0)
ORDER BY TABLE_NAME

Answer 7

Otra forma (para 2000 / 2005/2012/2014):

IF ((SELECT OBJECTPROPERTY( OBJECT_ID(N''table_name_here''), ''TableHasIdentity'')) = 1)
PRINT ''Yes''
ELSE
PRINT ''No''

NOTA: schema.table debe ser schema.table , a menos que el esquema sea dbo .

Answer 8

Otra forma potencial de hacer esto para SQL Server, que tiene menos confianza en las tablas del sistema (que están sujetas a cambios, versión a versión) es usar las vistas INFORMATION_SCHEMA:

select COLUMN_NAME, TABLE_NAME
from INFORMATION_SCHEMA.COLUMNS
where COLUMNPROPERTY(object_id(TABLE_SCHEMA+''.''+TABLE_NAME), COLUMN_NAME, ''IsIdentity'') = 1
order by TABLE_NAME

Answer 9

Utilizar esta :

DECLARE @Table_Name VARCHAR(100)
DECLARE @Column_Name VARCHAR(100)
SET @Table_Name = ''''
SET @Column_Name = ''''
SELECT RowNumber = ROW_NUMBER() OVER ( PARTITION BY T.[Name] ORDER BY T.[Name], C.column_id ) ,
SCHEMA_NAME(T.schema_id) AS SchemaName ,
T.[Name] AS Table_Name ,
C.[Name] AS Field_Name ,
sysType.name ,
C.max_length ,
C.is_nullable ,
C.is_identity ,
C.scale ,
C.precision
FROM Sys.Tables AS T
LEFT JOIN Sys.Columns AS C ON ( T.[Object_Id] = C.[Object_Id] )
LEFT JOIN sys.types AS sysType ON ( C.user_type_id = sysType.user_type_id )
WHERE ( Type = ''U'' )
AND ( C.Name LIKE ''%'' + @Column_Name + ''%'' )
AND ( T.Name LIKE ''%'' + @Table_Name + ''%'' )
ORDER BY T.[Name] ,
C.column_id

Answer 10

aquí hay una versión funcional para MSSQL 2000. He modificado el código 2005 que se encuentra aquí: http://sqlfool.com/2011/01/identity-columns-are-you-nearing-the-limits/

/* Define how close we are to the value limit
before we start throwing up the red flag.
The higher the value, the closer to the limit. */
DECLARE @threshold DECIMAL(3,2);
SET @threshold = .85;
/* Create a temp table */
CREATE TABLE #identityStatus
(
database_name VARCHAR(128)
, table_name VARCHAR(128)
, column_name VARCHAR(128)
, data_type VARCHAR(128)
, last_value BIGINT
, max_value BIGINT
);
DECLARE @dbname sysname;
DECLARE @sql nvarchar(4000);
-- Use an cursor to iterate through the databases since in 2000 there''s no sp_MSForEachDB command...
DECLARE c cursor FAST_FORWARD FOR
SELECT
name
FROM
master.dbo.sysdatabases
WHERE
name NOT IN(''master'', ''model'', ''msdb'', ''tempdb'');
OPEN c;
FETCH NEXT FROM c INTO @dbname;
WHILE @@FETCH_STATUS = 0
BEGIN
SET @sql = N''Use ['' + @dbname + ''];
Insert Into #identityStatus
Select '''''' + @dbname + '''''' As [database_name]
, Object_Name(id.id) As [table_name]
, id.name As [column_name]
, t.name As [data_type]
, IDENT_CURRENT(Object_Name(id.id)) As [last_value]
, Case
When t.name = ''''tinyint'''' Then 255
When t.name = ''''smallint'''' Then 32767
When t.name = ''''int'''' Then 2147483647
When t.name = ''''bigint'''' Then 9223372036854775807
End As [max_value]
From
syscolumns As id
Join systypes As t On id.xtype = t.xtype
Where
id.colstat&1 = 1 -- this identifies the identity columns (as far as I know)
'';
EXECUTE sp_executesql @sql;
FETCH NEXT FROM c INTO @dbname;
END
CLOSE c;
DEALLOCATE c;
/* Retrieve our results and format it all prettily */
SELECT database_name
, table_name
, column_name
, data_type
, last_value
, CASE
WHEN last_value < 0 THEN 100
ELSE (1 - CAST(last_value AS FLOAT(4)) / max_value) * 100
END AS [percentLeft]
, CASE
WHEN CAST(last_value AS FLOAT(4)) / max_value >= @threshold
THEN ''warning: approaching max limit''
ELSE ''okay''
END AS [id_status]
FROM #identityStatus
ORDER BY percentLeft;
/* Clean up after ourselves */
DROP TABLE #identityStatus;

Answer 11

sys.columns.is_identity = 1

p.ej,

select o.name, c.name
from sys.objects o inner join sys.columns c on o.object_id = c.object_id
where c.is_identity = 1