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Find cid of customers who have ordered exactly the same set of products which are ordered by a customer with cid 'c0001'.

I am doing an assignment for school which involves writing queries. I am done except for I stumped on one question:

Find cid of customers who have ordered exactly the same set of products which are ordered by a customer with cid 'c0001'.

I can't seem to think of anyway to determine that the sets are the same. I will post the table definitions for the assignment and what I have so far. It is due on Tuesday so a fast response would be appreciated.
SELECT c.cid
FROM customers c, orders o, orderline ol
WHERE c.cid = o.cid AND o.oid = ol.oid AND ol.pid IN(SELECT DISTINCT ol.pid FROM customers c, orders o, orderline ol WHERE c.cid = 'c0001' AND c.cid = o.cid AND o.oid = ol.oid);
 
-- We will drop the tables in case they exist from previous runs
drop table orderline;
drop table orders;
drop table products;
drop table categories; 
drop table customers;
  
CREATE TABLE customers (
  cid           char(5),
  cname         char(15),
  phone         char(12),
  address       char(20),
  PRIMARY KEY (cid)
);
CREATE TABLE categories (
  catid         char(2),
  catname       char(12),
  PRIMARY KEY (catid)
);
CREATE TABLE products (
  pid           char(5),
  pname         char(15),
  unit          char(5),
  price         float,
  stock_qty     integer,
  catid         char(2),
  PRIMARY KEY (pid),
  FOREIGN KEY (catid) REFERENCES categories
);
CREATE TABLE orders (
  oid           char(5),
  cid           char(5),
  odate         date,
  charge_amount float,
  PRIMARY KEY (oid),
  FOREIGN KEY (cid) REFERENCES customers
);
CREATE TABLE orderline (
  lno           integer,
  oid           char(5),
  pid           char(5),
  qty           integer, 
  price         float, 
  PRIMARY KEY (oid, lno),
  FOREIGN KEY (oid) REFERENCES orders ON DELETE CASCADE,
  FOREIGN KEY (pid) REFERENCES products
);

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bejhan
Asked:
bejhan
  • 5
  • 2
1 Solution
 
sujith80Commented:
Guidelines for solving this problem.

1. Get the count at customer level.
2. Join this results with another subquery which gets the count of orders for customer 'c0001'.
3. filter results with the count from step 1 matching with step 2.
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awking00Commented:
Answering homework questions can be tricky because Oracle often offers multiple ways of finding a solution and the responders may not be aware of the class topic. In your case, you need to identify two sets of results - those whose customer is c0001 and those who are not c0001 - and then compare them for equality. The comparison might take the form of a join, the use of something like the set operator intersect, or it may involve the use of a group by function with a "having" clause. Good luck!
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bejhanAuthor Commented:
sujith80, but the same count wouldn't necessarily mean the same set of items.
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bejhanAuthor Commented:
I am totally stumped on this:

I have made a queries that finds customers who have the same number of products as c0001 but  I don't know how to check if the products are all the same.
SELECT b.cid
FROM (SELECT c.cid, COUNT(DISTINCT ol.pid) AS Count
FROM customers c, orders o, orderline ol
WHERE c.cid = 'c0001' AND o.oid = ol.oid
GROUP BY c.cid) a, 
(SELECT c.cid, COUNT(DISTINCT ol.pid) AS Count
FROM customers c, orders o, orderline ol
WHERE c.cid = o.cid AND o.oid = ol.oid
GROUP BY c.cid) b
WHERE a.count = b.count;

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sujith80Commented:
follow this
SQL> select * from tbl1;
 
       CID PRODUCT                                                              
---------- --------------------                                                 
        10 prod1                                                                
        10 prod0                                                                
        20 prod2                                                                
        20 prod0                                                                
        40 prod1                                                                
        40 prod0                                                                
        50 prod5                                                                
        50 prod0                                                                
        60 prod1                                                                
        60 prod0                                                                
        70 prod0                                                                
 
11 rows selected.
 
SQL> select distinct cid
  2  from (
  3  select
  4   t.cid, t.product, X.cn, count(*) over (partition by t.cid) ct
  5  from
  6  tbl1 t,
  7  (select cid, product , count(*) over() cn from tbl1 where cid = 40) X
  8  where
  9  t.product = X.product
 10  )
 11  where cn = ct and cid <> 40 ;
 
       CID                                                                      
----------                                                                      
        10                                                                      
        60                                                                      
 
SQL> spool off

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bejhanAuthor Commented:
Okay I figured it out just in time :) I was looking at the question the wrong way, trying to include those who met requirements instead of excluding those who didn't.
SELECT cid
FROM customers
WHERE cid <> 'c0001'
MINUS
SELECT c.cid
FROM customers c, orders o, orderline ol 
WHERE c.cid <> 'c0001' AND c.cid = o.cid AND o.oid = ol.oid AND ol.pid NOT IN (SELECT ol.pid FROM orders o, orderline ol WHERE o.cid = 'c0001' AND o.oid = ol.oid) 
MINUS 
SELECT a.cid FROM (SELECT c.cid, COUNT(DISTINCT ol.pid) AS Count 
FROM customers c, orders o,  orderline ol WHERE c.cid <> 'c0001' AND c.cid = o.cid AND o.oid = ol.oid GROUP BY c.cid) a, (SELECT COUNT(DISTINCT ol.pid) AS Count FROM orders o,  orderline ol WHERE o.cid = 'c0001' AND o.oid = ol.oid GROUP BY o.cid) b 
WHERE a.count <> b.count;

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bejhanAuthor Commented:
I did not see that sujith80 had posted a solution, my apologies.
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bejhanAuthor Commented:
I did not see that sujith80 had posted a solution, my apologies.
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