如何在mysql中確定該用戶本周的滯納金

Question

所以在這里我有一個案例來確定這個用戶他每周是否延遲付款，以確定與付款日期和上一個逾期字段的延遲比較

我有這樣的樣本數據

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

希望生成的數據

 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

正如你看到的用戶名 bakrie 他延遲付款，因為在第二個付款日期支付的到期日超過了他付款的前一個記錄到期日

我試過這樣查詢但失敗了。 他這樣顯示數據不遲

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

我的查詢是這樣的`

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

請幫我！？！？！

Answer 1

我根據您對問題的描述來到了以下查詢。

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;

輸出：

+--------+--------+------------+--------------+----------+
|  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在線測試。