我如何创建一个外国的关键在SQL服务器？

Question

我从来没有"手工编码的"对象的创建代码SQL服务器和外键decleration似乎是之间的不同SQL服务器和Postgres。这里是我的sql迄今为止:

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，我得到这个错误：

Msg8139级别16状态0,9号线 数量引用列在 外键的不同之处数 引用列表 'question_bank'.

你能发现的错误？

Answer 1

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)
);

Answer 2

如果你只是想创建约束自己，可以使用改变表

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

我不会推荐该法提到通过萨拉Chipps内创建的，只是因为我宁名我自己的约束。

Answer 3

你也可以命名你的外国主要制约因素通过使用：

CONSTRAINT your_name_here FOREIGN KEY (question_exam_id) REFERENCES EXAMS (exam_id)

Answer 4

我喜欢AlexCuse的答案，但一些东西你应该注意到，只要你增加一个外国的关键制约因素是如何你想要的更新，引用列在行引用的表来对待，特别是如何你想要删除的排中引用的表格加以处理。

如果一个制约因素是创建了这样的：

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

..然后 更新或删除中引用的表会吹了一个错误，如果有一个相应的行中引用的表格。

可能的行为，但以我的经验，更多的通常是没有的。

如果你而不是创造这样的：

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

..然后再更新和删除在父母表将导致更新和删除相应的行中引用的表格。

(我不说默认应该改变的，默认犯错的谨慎，这是很好的。我只是说这件事情，一个人是谁创造constaints 始终应该注意.)

这是可以做到的，通过这种方式，当创建一个表，像这样：

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
)

Answer 5

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
);

-会的工作。Pehaps多一点建造直观?

Answer 6

如果你想建立两个表中列入的关系所使用的查询尝试如下：

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)

Answer 7

创建一个外国的关键任何表

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)

Answer 8

就像你一样，我通常不会创建外国键方面，但是，如果由于某种原因我需要剧本么做我通常创建使用ms sql服务器的管理工作室和前节省的随后改变，我选择表格的设计者|产生变化的脚本

Answer 9

这个脚本是关于创建表与外国的关键和我加入参照完整性的约束 sql服务器.

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
);

Answer 10

Necromancing.
实际上，这样做是正确的一点点棘手。

你首先需要检查的主键存在于所列要设置你的外国的关键基准。

在这个例子中，一个外国的关键在表T_ZO_SYS_Language_Forms创建，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