How do you determine what SQL Tables have an identity column programmatically
我想在SQL Server 2005中创建一个列列表，这些列在T-SQL中具有标识列及其对应的表。
结果将类似于：
表名，列名
对于SQL Server而言，这样做的另一种可能的方法是使用
INFORMATION_SCHEMA观看次数：
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| 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 |
sys.columns.is_identity = 1
例如
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| 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 |
另一种方式(对于2000 / 2005/2012/2014)：
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| IF ((SELECT OBJECTPROPERTY( OBJECT_ID(N'table_name_here'), 'TableHasIdentity')) = 1)
PRINT 'Yes'
ELSE
PRINT 'No' |
注意：table_name_here应该为schema.table，除非架构为dbo。
在SQL 2005中：
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| select object_name(object_id), name
from sys.columns
where is_identity = 1 |
此查询似乎可以解决问题：
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| 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') |
以下查询对我有用：
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| 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 |
基于吉列尔莫答案的不带身份列的表列表：
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| SELECT DISTINCT TABLE_NAME
FROM INFORMATION_SCHEMA.COLUMNS
WHERE (TABLE_SCHEMA = 'dbo') AND (OBJECTPROPERTY(OBJECT_ID(TABLE_NAME), 'TableHasIdentity') = 0)
ORDER BY TABLE_NAME |
这是适用于MSSQL 2000的工作版本。我修改了此处找到的2005代码：http://sqlfool.com/2011/01/identity-columns-are-you-nearing-the-limits/
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| /* 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; |
由于某种原因，sql server在不同的表中保存了一些标识列，因此对我有用的代码如下：
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| 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 |
这适用于SQL Server 2005、2008和2012。
我发现sys.identity_columns并不包含我所有带有标识列的表。
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| 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; |
在文档页面上，状态栏也可以使用。您还可以添加四个部分的标识符，它可以在不同的服务器上运行。
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| 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; |
来源：
https://msdn.microsoft.com/zh-CN/library/ms186816.aspx
使用此：
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| 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 |
这对于使用Sql Server 2008的我来说是有效的：
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| 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 |
我认为这适用于SQL 2000：
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| 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 |
|
