2013-04-16 43 views

回答

35
SELECT 
    [schema] = s.name, 
    [table] = t.name 
FROM sys.schemas AS s 
INNER JOIN sys.tables AS t 
    ON s.[schema_id] = t.[schema_id] 
WHERE EXISTS 
(
    SELECT 1 FROM sys.identity_columns 
    WHERE [object_id] = t.[object_id] 
); 
7
 select COLUMN_NAME, TABLE_NAME 
     from INFORMATION_SCHEMA.COLUMNS 
     where TABLE_SCHEMA = 'dbo' 
     and COLUMNPROPERTY(object_id(TABLE_NAME), COLUMN_NAME, 'IsIdentity') = 1 
     order by TABLE_NAME 
+2

也請提供某種解釋,同時回答問題.. – Lal

+0

請添加代碼的簡要說明。僅有代碼的答案(有時)很好,但代碼+解釋答案(總是)更好。 – Barranka

4

我喜歡這種方法,因爲它使用WHERE EXISTS或COLUMNPROPERTY呼叫的,而不是加入的。需要注意的是該集團是唯一必要的,如果你一)有表有一個以上的IDENTITY列和b)不想重複結果:

SELECT 
    SchemaName = s.name, 
    TableName = t.name 
FROM 
    sys.schemas AS s 
    INNER JOIN sys.tables AS t ON s.schema_id = t.schema_id 
    INNER JOIN sys.columns AS c ON t.object_id = c.object_id 
    INNER JOIN sys.identity_columns AS ic on c.object_id = ic.object_id AND c.column_id = ic.column_id 
GROUP BY 
    s.name, 
    t.name 
ORDER BY 
    s.name, 
    t.name; 
-1

選擇OBJECT_NAME(OBJECT_ID)RROM sys.identity_columns 其中is_identity = 1;

0

下面的腳本就可以了:

SELECT a.name as TableName, 
    CASE WHEN b.name IS NULL 
    THEN 'No Identity Column' 
    ELSE b.name 
    END as IdentityColumnName 
FROM sys.tables a 
    LEFT JOIN sys.identity_columns b on a.object_id = b.object_id