移动到第二个窗体 [英] Moving to second form

问题描述

你们怎么创建这个表的第二种形式（主要是：{egbn，copy}）：

isbn AB-1234-X

authorID IC45

authorName I.Conn

标题最后的窗帘>
复制 2

分类侦测功能

userID xyz44

解决方案

关系在2NF iff

• 在1NF中，并且


• 每个非素数属性是
  ，取决于每个
  候选键的整体（不在
  的任何候选键的一部分） ）

唯一的候选键是{isbn，copy}。所以问题就成了三个问题。

1. 这个关系在1NF？


2. 非素数属性
   {authorID，authorName，title，
   classi fi cation，userID}只依赖于{isbn}
   ？


3. -prime属性
   {authorID，authorName，title，
   classi fi cation，userID} only only
   on {copy}？

b

。

然后我创建单独的表与
属性：isbn，autorID，autorName，
标题，分类和另一个
表，其属性为：isbn，copy，
userID。

是。在关系发言中，您用这两个投影替换了原始关系R.

• R ₁ = { isbn，复制，
  userid}


• R ₂ = { isbn
  作者姓名，标题，分类}

如果你这样做了，你应该可以再次创建R在{isbn}上加入R 1和R 2，

现在两个R< 1< / sub >和R 2 <2>在2NF中。 （我认为这是作业问题的要点。）您可能需要考虑R ₁ 和R ₂ 是否在

• 3NF


• BCNF


• 4NF


• 5NF

---

稍后。 。

非正式地说，一个关系在3NF iff

• 它在2NF，而


• 没有传递依赖。

当我说它是在2NF，我的意思是这个关系在2NF 和它不是已经在3NF，BCNF，4NF或5NF。

什么正常形式是R< 1>和< 2> 2< / sub>你会想解释你的推理，否则你的讲师有责任让你看起来很愚蠢。我们不想要这样

• R ₁ = { isbn，复制，
  userid}


• R ₂ = { isbn ，authorid，
  authorname，title，classification}

而且稍后。 。 。

R ₁ 在5NF。 R ₂ 在2NF中。

R ₂ 不在3NF中，因为 isbn和authorname。

• isbn-> authorid，


• authorid-> authorname

通过用这两个投影替换R ₂ 来删除此传递依赖关系（R < sub>和R ₄ ）。

• R ₁ = { isbn，copy ，userid}（5NF）


• R ₃ = { isbn ，authorid，title，classification}


• R ₄ = { authorid ，authorname}

我认为标题和分类之间不存在功能依赖关系。

Guys, how would you create second form of this table (primary key is: {isbn,copy}):
isbn AB-1234-X
authorID IC45
authorName I.Conn
title The final curtain
copy 2
classification Detectivefiction
userID xyz44

解决方案

A relation is in 2NF iff

• it's in 1NF, and
• every non-prime attribute is dependent on the whole of every candidate key (not on just part of any candidate key)

The only candidate key is {isbn, copy}. So the question becomes three questions.

1. Is this relation in 1NF?
2. Are any of the non-prime attributes {authorID, authorName, title, classification, userID} dependent only on {isbn}?
3. Are any of the non-prime attributes {authorID, authorName, title, classification, userID} dependent only on {copy}?

What do you think?

---

Later . . .

Then I'm creating separate table with attributes: isbn, autorID, autorName, title, Classification and another table with attributes: isbn, copy, userID.

Yes. In "relational speak", you replaced the original relation R with these two projections.

• R₁ = {isbn, copy, userid}
• R₂ = {isbn, authorid, authorname, title, classification}

If you've done that correctly, you should be able to create R again by joining R₁ and R₂ on {isbn}.

Now both R₁ and R₂ are in 2NF. (I think that was the point of the homework question.) You might want to consider whether R₁ and R₂ are in

• 3NF
• BCNF
• 4NF
• 5NF

---

Still later . . .

Speaking informally, a relation is in 3NF iff

• it's in 2NF, and
• there are no transitive dependencies.

When I say "it's in 2NF", I mean the relation in question is in 2NF and it's not already in 3NF, BCNF, 4NF, or 5NF.

What normal form are R₁ and R₂ in? You'll want to explain your reasoning, otherwise your lecturer is liable to make you look foolish. And we don't want that.

• R₁ = {isbn, copy, userid}
• R₂ = {isbn, authorid, authorname, title, classification}

And still later . . .

R₁ is in 5NF. R₂ is in 2NF.

R₂ isn't in 3NF, because there's a transitive dependency between "isbn" and "authorname".

• isbn->authorid, and
• authorid->authorname

Remove this transitive dependency by replacing R₂ with these two projections (R₃ and R₄).

• R₁ = {isbn, copy, userid} (5NF)
• R₃ = {isbn, authorid, title, classification}
• R₄ = {authorid, authorname}

I don't think there's a functional dependency between title and classification.

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