Oracle SQL：表联接

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 ;