254

我从来没有为 SQL Server“手工编码”过对象创建代码，而且 SQL Server 和 Postgres 之间的外键解密似乎不同。到目前为止，这是我的 sql：

drop table exams;
drop table question_bank;
drop table anwser_bank;

create table exams
(
    exam_id uniqueidentifier primary key,
    exam_name varchar(50),
);
create table question_bank
(
    question_id uniqueidentifier primary key,
    question_exam_id uniqueidentifier not null,
    question_text varchar(1024) not null,
    question_point_value decimal,
    constraint question_exam_id foreign key references exams(exam_id)
);
create table anwser_bank
(
    anwser_id           uniqueidentifier primary key,
    anwser_question_id  uniqueidentifier,
    anwser_text         varchar(1024),
    anwser_is_correct   bit
);

当我运行查询时，我收到此错误：

消息 8139，级别 16，状态 0，第 9 行外键中的引用列数与表“question_bank”中引用的列数不同。

你能发现错误吗？

4

11 回答 11

338

如果你只想自己创建约束，你可以使用 ALTER TABLE

alter table MyTable
add constraint MyTable_MyColumn_FK FOREIGN KEY ( MyColumn ) references MyOtherTable(PKColumn)

我不推荐 Sara Chipps 提到的用于内联创建的语法，因为我宁愿命名我自己的约束。

于 2008-09-07T23:14:56.343 回答
206 
create table question_bank
(
    question_id uniqueidentifier primary key,
    question_exam_id uniqueidentifier not null,
    question_text varchar(1024) not null,
    question_point_value decimal,
    constraint fk_questionbank_exams foreign key (question_exam_id) references exams (exam_id)
);
于 2008-09-07T20:57:59.120 回答
70

您还可以使用以下方法命名外键约束：

CONSTRAINT your_name_here FOREIGN KEY (question_exam_id) REFERENCES EXAMS (exam_id)
于 2008-09-07T22:12:57.443 回答
32

我喜欢 AlexCuse 的回答，但是当您添加外键约束时，您应该注意的是，您希望如何更新被引用表的行中的被引用列，尤其是如何删除被引用表中的行要处理的表。

如果像这样创建约束：

alter table MyTable
add constraint MyTable_MyColumn_FK FOREIGN KEY ( MyColumn ) 
references MyOtherTable(PKColumn)

.. 然后，如果引用表中有相应的行，则引用表中的更新或删除将报错。

这可能是您想要的行为，但根据我的经验，通常不是。

如果您改为这样创建它：

alter table MyTable
add constraint MyTable_MyColumn_FK FOREIGN KEY ( MyColumn ) 
references MyOtherTable(PKColumn)
on update cascade 
on delete cascade

..then 父表中的更新和删除将导致引用表中相应行的更新和删除。

（我并不是说应该更改默认值，默认错误是谨慎行事，这很好。我只是说创建约束的人应该始终注意这一点。）

顺便说一句，这可以在创建表时完成，如下所示：

create table ProductCategories (
  Id           int identity primary key,
  ProductId    int references Products(Id)
               on update cascade on delete cascade
  CategoryId   int references Categories(Id) 
               on update cascade on delete cascade
)
于 2014-06-25T17:34:02.910 回答
14 
create table question_bank
(
    question_id uniqueidentifier primary key,
    question_exam_id uniqueidentifier not null constraint fk_exam_id foreign key references exams(exam_id),
    question_text varchar(1024) not null,
    question_point_value decimal
);

——这也行。也许更直观的构造？

于 2011-06-09T18:15:54.957 回答
8

在任何表上创建外键

ALTER TABLE [SCHEMA].[TABLENAME] ADD FOREIGN KEY (COLUMNNAME) REFERENCES [TABLENAME](COLUMNNAME)
EXAMPLE
ALTER TABLE [dbo].[UserMaster] ADD FOREIGN KEY (City_Id) REFERENCES [dbo].[CityMaster](City_Id)
于 2016-05-26T05:14:52.287 回答
8

如果您想通过使用查询将两个表的列创建为关系，请尝试以下操作：

Alter table Foreign_Key_Table_name add constraint 
Foreign_Key_Table_name_Columnname_FK
Foreign Key (Column_name) references 
Another_Table_name(Another_Table_Column_name)
于 2017-07-31T18:05:40.793 回答
5

像你一样，我通常不手动创建外键，但如果出于某种原因我需要脚本来执行此操作，我通常使用 ms sql server management studio 创建它，然后在保存更改之前，我选择 Table Designer | 生成更改脚本

于 2008-09-09T16:59:47.850 回答
5

该脚本是关于使用外键创建表的，并且我添加了引用完整性约束sql-server。

create table exams
(  
    exam_id int primary key,
    exam_name varchar(50),
);

create table question_bank 
(
    question_id int primary key,
    question_exam_id int not null,
    question_text varchar(1024) not null,
    question_point_value decimal,
    constraint question_exam_id_fk
       foreign key references exams(exam_id)
               ON DELETE CASCADE
);
于 2017-05-27T23:00:00.097 回答
3

死灵术。
实际上，正确地执行此操作有点棘手。

您首先需要检查要设置外键引用的列的主键是否存在。

在此示例中，创建表 T_ZO_SYS_Language_Forms 上的外键，引用 dbo.T_SYS_Language_Forms.LANG_UID

-- First, chech if the table exists...
IF 0 < (
    SELECT COUNT(*) FROM INFORMATION_SCHEMA.TABLES 
    WHERE TABLE_TYPE = 'BASE TABLE'
    AND TABLE_SCHEMA = 'dbo'
    AND TABLE_NAME = 'T_SYS_Language_Forms'
)
BEGIN
    -- Check for NULL values in the primary-key column
    IF 0 = (SELECT COUNT(*) FROM T_SYS_Language_Forms WHERE LANG_UID IS NULL)
    BEGIN
        ALTER TABLE T_SYS_Language_Forms ALTER COLUMN LANG_UID uniqueidentifier NOT NULL 

        -- No, don't drop, FK references might already exist...
        -- Drop PK if exists 
        -- ALTER TABLE T_SYS_Language_Forms DROP CONSTRAINT pk_constraint_name 
        --DECLARE @pkDropCommand nvarchar(1000) 
        --SET @pkDropCommand = N'ALTER TABLE T_SYS_Language_Forms DROP CONSTRAINT ' + QUOTENAME((SELECT CONSTRAINT_NAME FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS 
        --WHERE CONSTRAINT_TYPE = 'PRIMARY KEY' 
        --AND TABLE_SCHEMA = 'dbo' 
        --AND TABLE_NAME = 'T_SYS_Language_Forms' 
        ----AND CONSTRAINT_NAME = 'PK_T_SYS_Language_Forms' 
        --))
        ---- PRINT @pkDropCommand 
        --EXECUTE(@pkDropCommand) 

        -- Instead do
        -- EXEC sp_rename 'dbo.T_SYS_Language_Forms.PK_T_SYS_Language_Forms1234565', 'PK_T_SYS_Language_Forms';


        -- Check if they keys are unique (it is very possible they might not be) 
        IF 1 >= (SELECT TOP 1 COUNT(*) AS cnt FROM T_SYS_Language_Forms GROUP BY LANG_UID ORDER BY cnt DESC)
        BEGIN

            -- If no Primary key for this table
            IF 0 =  
            (
                SELECT COUNT(*) FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS 
                WHERE CONSTRAINT_TYPE = 'PRIMARY KEY' 
                AND TABLE_SCHEMA = 'dbo' 
                AND TABLE_NAME = 'T_SYS_Language_Forms' 
                -- AND CONSTRAINT_NAME = 'PK_T_SYS_Language_Forms' 
            )
                ALTER TABLE T_SYS_Language_Forms ADD CONSTRAINT PK_T_SYS_Language_Forms PRIMARY KEY CLUSTERED (LANG_UID ASC)
            ;

            -- Adding foreign key
            IF 0 = (SELECT COUNT(*) FROM INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS WHERE CONSTRAINT_NAME = 'FK_T_ZO_SYS_Language_Forms_T_SYS_Language_Forms') 
                ALTER TABLE T_ZO_SYS_Language_Forms WITH NOCHECK ADD CONSTRAINT FK_T_ZO_SYS_Language_Forms_T_SYS_Language_Forms FOREIGN KEY(ZOLANG_LANG_UID) REFERENCES T_SYS_Language_Forms(LANG_UID); 
        END -- End uniqueness check
        ELSE
            PRINT 'FSCK, this column has duplicate keys, and can thus not be changed to primary key...' 
    END -- End NULL check
    ELSE
        PRINT 'FSCK, need to figure out how to update NULL value(s)...' 
END
于 2015-10-23T10:02:18.763 回答
2

我总是使用这种语法在 2 个表之间创建外键约束

Alter Table ForeignKeyTable
Add constraint `ForeignKeyTable_ForeignKeyColumn_FK`
`Foreign key (ForeignKeyColumn)` references `PrimaryKeyTable (PrimaryKeyColumn)`

IE

Alter Table tblEmployee
Add constraint tblEmployee_DepartmentID_FK
foreign key (DepartmentID) references tblDepartment (ID)
于 2019-08-23T09:04:47.263 回答