Question

因此，在这里我有一个案例来确定该用户每周是否迟到他的付款，以确定与付款日期和先前的过期字段的迟到比较

我有这样的样本数据

Name   to be paid    DATE PAID       OVERDUE DATE
Bakrie  195000       2019-07-01       2019-07-08
Rocky   195000       2019-07-01       2019-07-08
Bakrie  195000       2019-07-15       2019-07-22
Bakrie  195000       2019-07-29       2019-08-05
Bakrie  195000       2019-08-05       2019-08-12
Febri   130000       2019-06-25       2019-07-02

希望生成的数据

 Name   to be paid    DATE PAID       OVERDUE DATE    sign
Bakrie  195000       2019-07-01       2019-07-08      NOT LATE
Rocky   195000       2019-07-01       2019-07-08      NOT LATE
Bakrie  195000       2019-07-15       2019-07-22      LATE
Bakrie  195000       2019-07-29       2019-08-05      Late
Bakrie  195000       2019-08-05       2019-08-12      not late
Febri   130000       2019-06-25       2019-07-02     not late

您看到 用户名bakrie他延迟付款，因为在第二个付款日期中付款已超过了他的付款的先前记录到期日期中的到期日期

我已经尝试过这样的查询，但是失败了。他这样显示数据不晚

Bakrie  195000  2019-07-01  2019-07-08  NOT LATE
Bakrie  195000  2019-07-01  2019-07-08  NOT LATE
Bakrie  195000  2019-07-01  2019-07-08  NOT LATE
Bakrie  195000  2019-07-01  2019-07-08  NOT LATE
Rocky   195000  2019-07-01  2019-07-08  NOT LATE
Bakrie  195000  2019-07-15  2019-07-22  LATE

我的查询是这样的

  select trackku.*, if(date paid > overdue_paid,'LATE','NOT LATE') from trackku

请帮助我！？！？！

Answer 1

我根据您对问题的描述来到了以下查询。

select
  a.name, a.paid, a.date_paid, a.overdue_date,
  case
    when max(b.overdue_date) is null or
         max(b.overdue_date) >= a.date_paid then
      'not late'
    else
      'late'
  end as sign
from trackku as a
left join trackku as b
  on a.name = b.name and a.date_paid > b.date_paid
group by a.name, a.paid, a.date_paid, a.overdue_date;

输出：

+--------+--------+------------+--------------+----------+
|  name  |  paid  | date_paid  | overdue_date |   sign   |
+--------+--------+------------+--------------+----------+
| Bakrie | 195000 | 2019-07-01 | 2019-07-08   | not late |
| Bakrie | 195000 | 2019-07-15 | 2019-07-22   | late     |
| Bakrie | 195000 | 2019-07-29 | 2019-08-05   | late     |
| Bakrie | 195000 | 2019-08-05 | 2019-08-12   | not late |
| Febri  | 130000 | 2019-06-25 | 2019-07-02   | not late |
| Rocky  | 195000 | 2019-07-01 | 2019-07-08   | not late |
+--------+--------+------------+--------------+----------+

使用SQL Fiddle在线进行测试。

如何在mysql中确定此用户本周的滞纳金

1 个答案: