Comment déterminer quelles tables SQL ont une colonne d'identité par programme

Question

Je veux créer une liste de colonnes dans SQL Server 2005 qui ont des colonnes d'identité et leur table correspondante dans T-SQL.

Les résultats seraient quelque chose comme:

TableName, NomColonne

Answer 1

Une autre façon possible de le faire pour SQL Server, qui a moins de dépendance sur les tables système (qui sont sujets à changement, une version à) est d'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 

Answer 2

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 

Answer 3

Dans SQL 2005:

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

Answer 4

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 

Answer 5

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') 

Answer 6

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.

Answer 7

est ici 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; 

Answer 8

Cela a fonctionné pour moi en utilisant 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 9

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 

Answer 10

Liste des tables 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 

Answer 11

Cela a fonctionné pour SQL Server 2005, 2008 et 2012. Je trouvé que le sys.identity_columns ne contenait 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 d'état peut également être utilisée. Vous pouvez également 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