简单地说，实体之间的“非识别”关系是ER-Model的一部分，在设计ER-Diagram时可以在Microsoft Visio中使用。这需要在类型为“0或大于0”或“0或1”的实体之间强制执行基数。注意基数中的“0”，而不是“一对多”中的“1”。

现在，一个非识别关系的例子，其中基数可能是“零”(非识别)，当我们说一个实体- a中的记录/对象“可能”或“可能”有一个值作为对另一个实体- b中的记录/s的引用。

由于实体a的一条记录有可能向其他实体b的记录标识自己，因此实体b中应该有一列具有实体b记录的标识值。如果实体a中没有记录标识实体b中的记录或对象，则此列可能为“Null”。

在面向对象(现实世界)范式中，b类对象的存在不一定依赖于a类对象(强耦合)，这意味着b类与a类是松散耦合的，因此a类可以“包含”a类对象(Containment)，而不是b类对象必须有a类对象(Composition)的概念，才能创建b类对象。

从SQL Query的角度来看，可以查询实体b中所有“非空”的外键为实体b保留的记录。这将为实体a中的行带来具有特定对应值的所有记录，或者所有具有Null值的记录将是实体b中实体a中没有任何记录的记录。

简单回答:是的，它可以是NULL或重复。

我想解释为什么外键可能需要为空，或者可能需要唯一或不唯一。首先记住，外键只是要求该字段中的值必须首先存在于另一个表(父表)中。这就是所有FK的定义。Null根据定义不是一个值。Null意味着我们还不知道值是什么。

Let me give you a real life example. Suppose you have a database that stores sales proposals. Suppose further that each proposal only has one sales person assigned and one client. So your proposal table would have two foreign keys, one with the client ID and one with the sales rep ID. However, at the time the record is created, a sales rep is not always assigned (because no one is free to work on it yet), so the client ID is filled in but the sales rep ID might be null. In other words, usually you need the ability to have a null FK when you may not know its value at the time the data is entered, but you do know other values in the table that need to be entered. To allow nulls in an FK generally all you have to do is allow nulls on the field that has the FK. The null value is separate from the idea of it being an FK.

Whether it is unique or not unique relates to whether the table has a one-one or a one-many relationship to the parent table. Now if you have a one-one relationship, it is possible that you could have the data all in one table, but if the table is getting too wide or if the data is on a different topic (the employee - insurance example @tbone gave for instance), then you want separate tables with a FK. You would then want to make this FK either also the PK (which guarantees uniqueness) or put a unique constraint on it.

大多数FK是一对多关系，这就是你从FK中得到的，而不需要在字段上增加进一步的约束。例如，你有一个订单表和订单明细表。如果客户一次订购10件商品，那么他有一个订单和10个订单详细记录，其中包含与FK相同的orderID。

我认为一个表的外键也是另一个表的主键。所以它不允许空值。所以外键没有空值的问题。

Yes foreign key can be null as told above by senior programmers... I would add another scenario where Foreign key will required to be null.... suppose we have tables comments, Pictures and Videos in an application which allows comments on pictures and videos. In comments table we can have two Foreign Keys PicturesId, and VideosId along with the primary Key CommentId. So when you comment on a video only VideosId would be required and pictureId would be null... and if you comment on a picture only PictureId would be required and VideosId would be null...

外键可以为NULL吗?

现有的答案集中在单列场景。如果我们考虑多列外键，我们使用SQL标准中定义的MATCH [SIMPLE | PARTIAL | FULL]子句有更多的选项:

PostgreSQL-CREATE TABLE A value inserted into the referencing column(s) is matched against the values of the referenced table and referenced columns using the given match type. There are three match types: MATCH FULL, MATCH PARTIAL, and MATCH SIMPLE (which is the default). MATCH FULL will not allow one column of a multicolumn foreign key to be null unless all foreign key columns are null; if they are all null, the row is not required to have a match in the referenced table. MATCH SIMPLE allows any of the foreign key columns to be null; if any of them are null, the row is not required to have a match in the referenced table. MATCH PARTIAL is not yet implemented. (Of course, NOT NULL constraints can be applied to the referencing column(s) to prevent these cases from arising.)

例子:

CREATE TABLE A(a VARCHAR(10), b VARCHAR(10), d DATE , UNIQUE(a,b));
INSERT INTO A(a, b, d) 
VALUES (NULL, NULL, NOW()),('a', NULL, NOW()),(NULL, 'b', NOW()),('c', 'b', NOW());

CREATE TABLE B(id INT PRIMARY KEY, ref_a VARCHAR(10), ref_b VARCHAR(10));

-- MATCH SIMPLE - default behaviour nulls are allowed
ALTER TABLE B ADD CONSTRAINT B_Fk FOREIGN KEY (ref_a, ref_b) 
REFERENCES A(a,b) MATCH SIMPLE;

INSERT INTO B(id, ref_a, ref_b) VALUES (1, NULL, 'b');  

-- (NULL/'x') 'x' value does not exists in A table, but insert is valid
INSERT INTO B(id, ref_a, ref_b) VALUES (2, NULL, 'x');  

ALTER TABLE B DROP CONSTRAINT IF EXISTS B_Fk; -- cleanup

-- MATCH PARTIAL - not implemented
ALTER TABLE B ADD CONSTRAINT B_Fk FOREIGN KEY (ref_a, ref_b) 
REFERENCES A(a,b) MATCH PARTIAL;
-- ERROR:  MATCH PARTIAL not yet implemented

DELETE FROM B; ALTER TABLE B DROP CONSTRAINT IF EXISTS B_Fk; -- cleanup

-- MATCH FULL nulls are not allowed
ALTER TABLE B ADD CONSTRAINT B_Fk FOREIGN KEY (ref_a, ref_b) 
REFERENCES A(a,b) MATCH FULL;

-- FK is defined, inserting NULL as part of FK
INSERT INTO B(id, ref_a, ref_b) VALUES (1, NULL, 'b');
-- ERROR:  MATCH FULL does not allow mixing of null and nonnull key values.

-- FK is defined, inserting all NULLs - valid
INSERT INTO B(id, ref_a, ref_b) VALUES (1, NULL, NULL);

db < > fiddle演示

简单地说，实体之间的“非识别”关系是ER-Model的一部分，在设计ER-Diagram时可以在Microsoft Visio中使用。这需要在类型为“0或大于0”或“0或1”的实体之间强制执行基数。注意基数中的“0”，而不是“一对多”中的“1”。

现在，一个非识别关系的例子，其中基数可能是“零”(非识别)，当我们说一个实体- a中的记录/对象“可能”或“可能”有一个值作为对另一个实体- b中的记录/s的引用。

由于实体a的一条记录有可能向其他实体b的记录标识自己，因此实体b中应该有一列具有实体b记录的标识值。如果实体a中没有记录标识实体b中的记录或对象，则此列可能为“Null”。

在面向对象(现实世界)范式中，b类对象的存在不一定依赖于a类对象(强耦合)，这意味着b类与a类是松散耦合的，因此a类可以“包含”a类对象(Containment)，而不是b类对象必须有a类对象(Composition)的概念，才能创建b类对象。

从SQL Query的角度来看，可以查询实体b中所有“非空”的外键为实体b保留的记录。这将为实体a中的行带来具有特定对应值的所有记录，或者所有具有Null值的记录将是实体b中实体a中没有任何记录的记录。