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对于 Oracle 数据库，假设我这里有两个表（结构相似，但数据量更大）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（按付款日期获取每个客户订单的前两个付款金额和日期）：

客户编号 | pay_amount1 | pay_date1 |pay_amount2 | pay_date2

我想要的样本报告

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

任何人都可以做出正确有效的查询吗？提前谢谢。

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3 回答 3

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首先，您需要正确设置创建脚本。;终止语句而不是语句中的行。其次，varchar2数据类型需要一个长度规范：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是数字。这是两个不同的东西。

所以这会产生以下脚本：

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，不指定INSERT语句中的列是不好的编码习惯。它应该insert into customer (cust_no, name, zip) values(1,'Tom','90001');代替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，我使用了外部联接来确保返回每个客户，即使没有或只有一次付款。

这是一个带有所有更正和查询的 SQLFiddle：http ://sqlfiddle.com/#!4/74349/3

于 2013-10-10T16:57:25.477 回答
1 
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;

将在 2 个单独的行中显示您需要的每个客户的 2 条记录。如果您更喜欢将其放在同一行中，请使用以下查询：

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;
于 2013-10-10T17:17:13.503 回答
0

分析功能ROW_NUMBER可以帮助您为每次付款提供一个数字：

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

使用这个我想我们可以得到你正在寻找的东西，比如：

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 ;
于 2013-10-10T16:57:26.087 回答