22

我目前正在使用...

select Table_Name, Column_name, data_type, is_Nullable
from information_Schema.Columns

...确定有关给定数据库中列的信息,以生成数据访问层。

我可以从哪里检索有关这些列是否是其表主键中的参与者的信息?

4

4 回答 4

43

这是一种方法(将“keycol”替换为您要搜索的列名):

SELECT  K.TABLE_NAME ,
    K.COLUMN_NAME ,
    K.CONSTRAINT_NAME
FROM    INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS C
        JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS K ON C.TABLE_NAME = K.TABLE_NAME
                                                         AND C.CONSTRAINT_CATALOG = K.CONSTRAINT_CATALOG
                                                         AND C.CONSTRAINT_SCHEMA = K.CONSTRAINT_SCHEMA
                                                         AND C.CONSTRAINT_NAME = K.CONSTRAINT_NAME
WHERE   C.CONSTRAINT_TYPE = 'PRIMARY KEY'
        AND K.COLUMN_NAME = 'keycol';
于 2008-10-21T15:08:24.697 回答
7

同样,以下将为您提供有关所有表及其键的信息,而不是有关特定列的信息。这样,您可以确保您拥有所有感兴趣的列并知道它们参与的内容。为了查看所有键(主键、外键、唯一键),请注释 WHERE 子句。

SELECT K.TABLE_NAME, C.CONSTRAINT_TYPE, K.COLUMN_NAME, K.CONSTRAINT_NAME
FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS C
JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS K
ON C.TABLE_NAME = K.TABLE_NAME
AND C.CONSTRAINT_CATALOG = K.CONSTRAINT_CATALOG
AND C.CONSTRAINT_SCHEMA = K.CONSTRAINT_SCHEMA
AND C.CONSTRAINT_NAME = K.CONSTRAINT_NAME
WHERE C.CONSTRAINT_TYPE = 'PRIMARY KEY'
ORDER BY K.TABLE_NAME, C.CONSTRAINT_TYPE, K.CONSTRAINT_NAME
于 2008-10-21T15:17:22.433 回答
3

根据您的需要,使用 INFORMATION_SCHEMA.COLUMNS 和 INFORMATION_SCHEMA.KEY_COLUMN_USAGE 进行完全外部联接。在 select 语句中,从 INFORMATION_SCHEMA.KEY_COLUMN_USAGE 添加 CONSTRAINT_NAME 列,这将为您提供 null 或键名。

select C.Table_Name, C.Column_name, data_type, is_Nullable, U.CONSTRAINT_NAME
from information_Schema.Columns C FULL OUTER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE U ON C.COLUMN_NAME = U.COLUMN_NAME
WHERE C.TABLE_NAME=@TABLENAME
于 2014-11-05T16:05:45.583 回答
0

此查询返回列是主键。

SELECT  col.COLUMN_NAME ,
        col.DATA_TYPE ,
        col.CHARACTER_MAXIMUM_LENGTH ln ,
        CAST(ISNULL(j.is_primary, 0) AS BIT) is_primary
FROM    INFORMATION_SCHEMA.COLUMNS col
        LEFT JOIN ( SELECT  K.TABLE_NAME ,
                            K.COLUMN_NAME ,
                            CASE WHEN K.CONSTRAINT_NAME IS NULL THEN 0
                                 WHEN K.CONSTRAINT_NAME IS NOT NULL THEN 1
                            END is_primary
                    FROM    INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS C
                            JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS K ON C.TABLE_NAME = K.TABLE_NAME
                                                              AND C.CONSTRAINT_CATALOG = K.CONSTRAINT_CATALOG
                                                              AND C.CONSTRAINT_SCHEMA = K.CONSTRAINT_SCHEMA
                                                              AND C.CONSTRAINT_NAME = K.CONSTRAINT_NAME
                    WHERE   C.CONSTRAINT_TYPE = 'PRIMARY KEY'
                            AND C.TABLE_NAME = 'tablename'
                  ) j ON col.COLUMN_NAME = j.COLUMN_NAME
WHERE   col.TABLE_NAME = 'tablename'
于 2019-06-06T10:07:56.623 回答