Oracle SQL : table Joining - sql

Question

For Oracle Database, suppose I have two tables here (Similar structure, but much larger amount of data) Definition below:
create table payments(
record_no INTEGER;
cust_no INTEGER;
amount NUMBER;
date_entered DATE;
);
insert into payments values(1,3,34.5,sysdate-1);
insert into payments values(2,2,34.5,sysdate-2);
insert into payments values(3,3,34.5,sysdate-18/1440);
insert into payments values(4,1,34.5,sysdate-1);
insert into payments values(5,2,34.5,sysdate-2/24);
insert into payments values(6,3,34.5,sysdate-56/1440);
insert into payments values(7,4,34.5,sysdate-2);
insert into payments values(8,2,34.5,sysdate-1);
create table customer(
cust_no INTEGER;
name VARCHAR2;
zip VARCHAR2;
);
insert into customer values(1,'Tom',90001);
insert into customer values(2,'Bob',90001);
insert into customer values(3,'Jack',90001);
insert into customer values(4,'Jay',90001);
Now I want to generate a report with those columns (Get the first two payment amount and date for each customer order by paydate) :
Cust_no | pay_amount1 | pay_date1 |pay_amount2 | pay_date2
Sample report I want
CUST_NO PAYMENT1 PAYDATE1 PAYMENT2 PAYDATE2
1 34.5 October, 09 2013 0 null
2 34.5 October, 08 2013 34.5 October, 09 2013
3 34.5 October, 09 2013 34.5 October, 10 2013
4 34.5 October, 08 2013 0 null
Can anybody make a correct and efficient Query ? Thanks ahead.

Answer 1

First you need to get your creation script right. The ; terminates a statement not a line inside a statement. Secondly the varchar2 data types needs a length specification: name VARCHAR2(20) instead of name VARCHAR2. Also character literals need to be enclosed in single quotes. '90001' is a character literal, 90001 is a number. Those are two different things.
So this results in the following script:
create table payments(
record_no INTEGER,
cust_no INTEGER,
amount NUMBER,
date_entered DATE
);
insert into payments values(1,3,34.5,sysdate-1);
insert into payments values(2,2,34.5,sysdate-2);
insert into payments values(3,3,34.5,sysdate-18/1440);
insert into payments values(4,1,34.5,sysdate-1);
insert into payments values(5,2,34.5,sysdate-2/24);
insert into payments values(6,3,34.5,sysdate-56/1440);
insert into payments values(7,4,34.5,sysdate-2);
insert into payments values(8,2,34.5,sysdate-1);
create table customer(
cust_no INTEGER,
name VARCHAR2(20),
zip VARCHAR2(20)
);
insert into customer values(1,'Tom','90001');
insert into customer values(2,'Bob','90001');
insert into customer values(3,'Jack','90001');
insert into customer values(4,'Jay','90001');
Note that it's bad coding practice to not specify the columns in an INSERT statement. It should be insert into customer (cust_no, name, zip) values(1,'Tom','90001'); instead of insert into customer values(1,'Tom','90001');
Now for your query, the following should do you wnat you need:
with numbered_payments as (
select cust_no,
amount,
date_entered,
row_number() over (partition by cust_no order by date_entered) as rn
from payments
)
select c.cust_no,
c.name,
p1.amount as pay_amount1,
p1.date_entered as pay_date1,
p2.amount as pay_amount2,
p2.date_entered as pay_date2
from customer c
left join numbered_payments p1
on p1.cust_no = c.cust_no
and p1.rn = 1
left join numbered_payments p2
on p2.cust_no = c.cust_no
and p2.rn = 2;
Note that I used an outer join to ensure that every customer is returned even if there is no or only a single payment for it.
Here is an SQLFiddle with all corrections and the query: http://sqlfiddle.com/#!4/74349/3

Answer 2

SELECT * FROM (
SELECT
c.cust_no,
p.amount as payment,
p.date_entered as paydate,
ROW_NUMBER() OVER (PARTITION BY cust_no ORDER BY p.record_no ASC) AS rn
FROM customer c
JOIN payments p ON p.cust_no = c.cust_no
) t
WHERE
rn <= 2
ORDER BY cust_no, rn;
Will show the 2 records per client you need, in 2 separate lines. If you prefer having it in the same line, then use this query:
SELECT
cust_no,
payment1,
paydate1,
CASE WHEN nextcli <> cust_no THEN 0 ELSE payment2 END AS payment2,
CASE WHEN nextcli <> cust_no THEN SYSDATE ELSE paydate2 END AS paydate2
FROM (
SELECT
c.cust_no,
p.amount as payment1,
p.date_entered as paydate1,
ROW_NUMBER() OVER (PARTITION BY c.cust_no ORDER BY p.record_no ASC) AS rn,
LEAD(c.cust_no, 1, -1) OVER (ORDER BY c.cust_no ASC) as nextcli,
LEAD(p.amount, 1, 0) OVER (ORDER BY c.cust_no ASC) as payment2,
LEAD(p.date_entered, 1, NULL) OVER (ORDER BY c.cust_no ASC) as paydate2
FROM customer c
JOIN payments p ON p.cust_no = c.cust_no
) t
WHERE
rn <= 1
ORDER BY cust_no, rn;

Answer 3

The analytic function ROW_NUMBER can help you give a number to each payment :
select cust_no, amount, date_entered,
row_number() over(partition by cust_no order by date_entered ) rn
from payments ;
Using this I think we can get what you're looking for, something like :
With ordered_payments as (
select cust_no, amount, date_entered,
row_number() over(partition by cust_no order by date_entered ) rn
from payments)
select customer.cust_no, p1.amount, p1.date_entered, p2.amount, p2.date_entered
from customer left join ordered_payments p1
on customer.cust_no = p1.cust_no and p1.rn = 1
left join ordered_payments p2
on customer.cust_no = p2.cust_no and p2.rn = 2 ;