जवाबों:
SQL सर्वर के लिए ऐसा करने का एक और संभावित तरीका है, जिसमें सिस्टम तालिकाओं पर कम निर्भरता है (जो कि परिवर्तन, संस्करण से संस्करण के अधीन हैं) INFORMATION_SCHEMA विचारों का उपयोग करना है:
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 INFORMATION_SCHEMA.COLUMNSसादा तालिकाओं और विचारों के लिए infos शामिल है और मैं परिणाम सेट की पठनीयता के लिए TABLE_TYPE (INFORMATION_SCHEMA.TABLES में शामिल होने) का सुझाव देता हूं।
sys.columns.is_identity = 1
जैसे,
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 = 1SQL 2005 में:
select object_name(object_id), name
from sys.columns
where is_identity = 1यह प्रश्न चाल करने के लिए लगता है:
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')यहाँ MSSQL 2000 के लिए एक कार्यशील संस्करण है। मैंने यहाँ पाया 2005 कोड को संशोधित किया है: 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;गिलर्मो के उत्तर के आधार पर बिना पहचान पत्र के तालिकाओं की सूची :
SELECT DISTINCT TABLE_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE (TABLE_SCHEMA = 'dbo') AND (OBJECTPROPERTY(OBJECT_ID(TABLE_NAME), 'TableHasIdentity') = 0)
ORDER BY TABLE_NAMEमेरे लिए निम्नलिखित क्वेरी काम करती है:
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_NAMEमुझे लगता है कि यह SQL 2000 के लिए काम करता है:
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 = @ColumnNameइसने मेरे लिए Sql Server 2008 का उपयोग करके काम किया:
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;
GOइसे इस्तेमाल करो :
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_idइसने SQL Server 2005, 2008 और 2012 के लिए काम किया। मैंने पाया कि sys.identity_columns में पहचान कॉलम के साथ मेरे सभी टेबल नहीं थे।
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;प्रलेखन पृष्ठ को देखते हुए स्थिति कॉलम का भी उपयोग किया जा सकता है। इसके अलावा आप चार भाग पहचानकर्ता जोड़ सकते हैं और यह विभिन्न सर्वरों पर काम करेगा।
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;स्रोत: https://msdn.microsoft.com/en-us/library/ms186816.aspx
किसी कारण से sql सर्वर कुछ पहचान स्तंभों को विभिन्न तालिकाओं में सहेजता है, जो कोड मेरे लिए काम करता है, वह निम्नलिखित है:
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