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Comment déterminez-vous quelles tables SQL ont une colonne d'identité par programme

Je souhaite créer une liste de colonnes dans SQL Server 2005 comportant des colonnes d'identité et la table correspondante dans T-SQL.

Les résultats seraient quelque chose comme:

Nom de la table, Nom de la colonne

sql-servertsqlmetadataidentity-column
88
Gabe

Un autre moyen potentiel de le faire pour SQL Server, qui s'appuie moins sur les tables système (qui sont sujettes à modification, version à version), consiste à utiliser les vues 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
152
DaveCrawford

sys.columns.is_identity = 1

par exemple.,

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
44
Kevin Crumley

Une autre façon (pour 2000/2005/2012/2014):

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

REMARQUE: table_name_here doit être schema.table, sauf si le schéma est dbo.

26
Guillermo

En SQL 2005:

select object_name(object_id), name
from sys.columns
where is_identity = 1
5
Euro Micelli

Cette requête semble faire l'affaire:

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')
2
Manrico Corazzi

voici une version de travail pour MSSQL 2000. J'ai modifié le code 2005 trouvé ici: 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;
1
S.E.

Liste des tableaux sans colonne Identité basée sur Guillermo réponse:

SELECT DISTINCT TABLE_NAME
FROM            INFORMATION_SCHEMA.COLUMNS
WHERE        (TABLE_SCHEMA = 'dbo') AND (OBJECTPROPERTY(OBJECT_ID(TABLE_NAME), 'TableHasIdentity') = 0)
ORDER BY TABLE_NAME
1
Sergey

Cela fonctionnait pour SQL Server 2005, 2008 et 2012. J'ai constaté que les colonnes sys.identity_columns ne contenaient pas toutes mes tables avec des colonnes d'identité.

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 regardant la page de documentation, la colonne de statut peut également être utilisée. Vous pouvez aussi ajouter l'identifiant en quatre parties et cela fonctionnera sur différents serveurs.

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;

Source: https://msdn.Microsoft.com/en-us/library/ms186816.aspx

0
Nikolai Bielik

La requête suivante fonctionne pour moi:

select  TABLE_NAME tabla,COLUMN_NAME columna
from    INFORMATION_SCHEMA.COLUMNS
where   COLUMNPROPERTY(object_id(TABLE_SCHEMA+'.'+TABLE_NAME), COLUMN_NAME, 'IsIdentity') = 1
order by TABLE_NAME
0
Jorge Santos Neill

Cela a fonctionné pour moi avec 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
0
James Drinkard

Pour une raison quelconque, SQL Server sauvegarde certaines colonnes d'identité dans différentes tables. Le code qui fonctionne pour moi est le suivant:

select      TABLE_NAME tabla,COLUMN_NAME columna
from        INFORMATION_SCHEMA.COLUMNS
where       COLUMNPROPERTY(object_id(TABLE_SCHEMA+'.'+TABLE_NAME), COLUMN_NAME, 'IsIdentity') = 1
union all
select      o.name tabla, c.name columna
from        sys.objects o 
inner join  sys.columns c on o.object_id = c.object_id
where       c.is_identity = 1
0
Jorge Santos Neill

Utilisez ceci :

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
0
Ardalan Shahgholi

Je pense que cela fonctionne pour 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
0
Brian