Question

我在oracle数据库中有一个如下表：

s_num     eff_date     amount
1         07-OCT-18      110
1         06-OCT-18      50
2         07-OCT-18      200
2         06-OCT-18      150

我想找到今天和昨天的“金额”列之间的区别。结果必须如下所示

s_num    diff
1         60
2         50

我想做到这一点而无需将表自身连接起来。

Answer 1

您可以使用条件聚合

SELECT s_num,SUM (CASE 
                    WHEN eff_date >= TRUNC(sysdate) 
                       AND eff_date < TRUNC(sysdate) + 1 THEN amount --today
                    WHEN eff_date >= TRUNC(sysdate) - 1 
                       AND eff_date < TRUNC(sysdate) THEN - amount --yesterday
                END) 
    FROM   t 
    GROUP  BY s_num;

或LAG函数

SELECT s_num, 
       diff 
FROM   (SELECT s_num, 
               eff_date,
               amount - LAG(amount, 1, 0) 
                          OVER ( 
                            partition BY s_num 
                            ORDER BY eff_date ) AS diff 
        FROM   t) 
WHERE  eff_date >= trunc(sysdate) 
       AND eff_date < trunc(sysdate) + 1 ;

Demo

除了今天以外，第一个会打印NULL差额，第二个不会为其他日期返回任何行。

Answer 2

sql>set serveroutput on;
sql>create or replace function difference(n in number) return number as di 
    number(5);
 2  begin
 3  select ((select amount from oc where dob=trunc(sysdate) and snum=n)- 
    (select 
    amount from oc where dob=trunc(sysdate-1) and snum=n)) into di from dual;
 4  return di;
 5  end;
 6  /
 sql>select distinct snum,difference(snum) as diff from oc;

输出：

                   SNUM       DIFF
            ---------- ----------
                    2         50
                    1         60

在两天内计算表格的一列之间的差异

2 个答案: