从关系代数翻译成 SQL

Question

下面的练习要求我将关系代数转换为 SQL 代码。

我对关系代数还不是很熟悉，但我尝试在SQL中编写以下关系，但我认为我犯了一些错误。

    **>  [Customer × Product ]―[π{Cid, Name, Pid, Label}(Customer ⋈ Orders ⋈ line_item)]**

SELECT *  FROM Customer, Product  WHERE Cid, Name, Pid, Label NOT IN
(SELECT Cid, Name, Pid, Label FROM Customer NATURAL JOIN Orders
NATURAL JOIN line_item);

这个代数关系我真的不知道怎么处理:

**>  πName,Name2(σCid<Cid2 (πCid,Name,Pid (Customer ⋈ Orders ⋈ line_item)
⋈ βCid→Cid2,Name→Name2 (πCid,Name,Pid (Customer ⋈ Orders ⋈
line_item))))**

如果您能向我解释处理此类代数关系的推理过程，我们将不胜感激。

Answer 1

对于第一个查询，它看起来几乎是正确的，除了我不认为你可以用 NOT IN 给出多个列。我会使用 WHERE NOT EXISTS:

SELECT * FROM Customer c1, Product
WHERE NOT EXISTS 
  (SELECT Cid, Name, Pid, Label FROM Customer c2
  NATURAL JOIN Orders
  NATURAL JOIN line_item
  WHERE c1.Cid = c2.Cid); -- assuming Cid is a primary key of your customer table

对于第二部分，

πName,Name2(σCid<Cid2 (πCid,Name,Pid (Customer ⋈ Orders ⋈ line_item)
⋈ βCid→Cid2,Name→Name2 (πCid,Name,Pid (Customer ⋈ Orders ⋈
line_item))))

可以这么写

R1 = (πCid,Name,Pid (Customer ⋈ Orders ⋈ line_item))

R2 = (πCid,Name,Pid (Customer ⋈ Orders ⋈ line_item) ⋈ βCid→Cid2,Name→Name2 R1))

R3 = πName,Name2(σCid<Cid2 R2)

这将翻译：

R1 = (SELECT Cid, Name, Pid FROM Customer NATURAL JOIN Orders NATURAL JOIN line_item)

R2 = (SELECT Cid, Name, Pid FROM Customer NATURAL JOIN Orders NATURAL JOIN line_item NATURAL JOIN (SELECT Cid as Cid2, Name as Name2 FROM R1))

R3 = SELECT Name, Name2 FROM R2 WHERE Cid < Cid2

从关系代数翻译成 SQL

Translating from Relational Algebra to SQL

sql

relational-algebra