Question

对于Oracle数据库，假设我这里有两个表（结构相似，但数据量大得多）定义如下：

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);

现在我想生成一个包含这些列的报告（通过paydate获取每个客户订单的前两个付款金额和日期）：

Cust_no | pay_amount1 | pay_date1 | pay_amount2 | pay_date2

我想要的示例报告

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

任何人都可以提出正确有效的查询吗？ 非常感谢。

Answer 1

首先，您需要正确创建脚本。 ;终止语句而不是语句中的一行。其次，varchar2数据类型需要长度规范：name VARCHAR2(20)而不是name VARCHAR2。字符文字也需要用单引号括起来。 '90001'是字符文字，90001是一个数字。这是两件不同的事情。

所以这导致以下脚本：

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');

请注意，不在INSERT语句中指定列是不好的编码习惯。它应该是insert into customer (cust_no, name, zip) values(1,'Tom','90001');而不是insert into customer values(1,'Tom','90001');

现在，对于您的查询，您需要以下内容：

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;

请注意，我使用外部联接来确保即使没有或只有一次付款，也会退回每个客户。

这是一个包含所有更正和查询的SQLFiddle：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;

将显示您需要的每个客户端的2条记录，分为两行。如果您希望将它放在同一行中，请使用此查询：

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

分析函数ROW_NUMBER可以帮助您为每笔付款提供一个数字：

select cust_no, amount, date_entered,
   row_number() over(partition by cust_no order by date_entered ) rn
from payments ;

使用这个我认为我们可以得到你想要的东西，例如：

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 ;

Oracle SQL：表连接

3 个答案: