如何在mysql中确定该用户本周的滞纳金

Question

所以在这里我有一个案例来确定这个用户他每周是否延迟付款，以确定与付款日期和上一个逾期字段的延迟比较

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

正如你看到的用户名 bakrie 他延迟付款，因为在第二个付款日期支付的到期日超过了他付款的前一个记录到期日

我试过这样查询但失败了。 他这样显示数据不迟

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在线测试。