Mysql按日期顺序检测doublon

Question

我对此请求有疑问： SELECT COUNT() AS nbr_doublon, tel, dateippo FROM vici GROUP BY tel HAVING COUNT() > 1

我想得到这个结果：

^结果将仅包含已尝试日期的副本 请帮忙

-------------- 表格数据

id  tel            dateippo     status  lead_id 
1   11111111111     2017-01-02  ok      4   y
2   11111111111     2017-01-26  na      5   y
3   11111111111     2017-01-28  rep     4   n
4   22222222222     2017-01-10  ok      7   y

我想要的结果：

id  tel             dateippo    status  lead_id 
1   11111111111     2017-01-02  ok      4   
4   22222222222     2017-01-10  ok      7   

首先测试状态是否等于“ok”并按 dateippo 订购 否则按日期订购

求帮助

Answer 1

假设您想为每个电话获取具有最大 id 的行，请使用：

select
    t.*
from vici t
inner join (
    select tel, max(id) id
    from vici
    group by tel
) t2 on t.tel = t2.tel
and t.id = t2.id;

编辑 1：

如果它基于 dateippo，只需将 id 替换为 dateippo，如下所示：

select
    t.*
from vici t
inner join (
    select tel, max(dateippo) dateippo
    from vici
    group by tel
) t2 on t.tel = t2.tel
and t.dateippo = t2.dateippo;

编辑 2：

首先按状态 = 'ok' 排序，然后按日期排序，

select
    t.*
from vici t
inner join (
    select tel, max(id) id
    from vici
    group by tel
) t2 on t.tel = t2.tel
and t.id = t2.id
order by t.status <> 'ok', t.dateippo;

编辑 3：

select
    t.*
from vici t
inner join (
    select 
        tel,
        case when count(status = 'ok') > 1 then 
            max(case when status = 'ok' then dateippo end) 
        else max(dateippo) end dateippo
    from vici
    group by tel
) t2 on t.tel = t2.tel
and t.dateippo = t2.dateippo;
select * from vici;

或基于id：

select
    t.*
from vici t
inner join (
    select 
        tel,
        case when count(status = 'ok') > 1 then 
            max(case when status = 'ok' then id end) 
        else max(id) end id
    from vici
    group by tel
) t2 on t.tel = t2.tel
and t.id = t2.id;
select * from vici;

+------+-------------+------------+--------+---------+
| id   | tel         | dateippo   | status | lead_id |
+------+-------------+------------+--------+---------+
|    1 | 11111111111 | 2017-01-02 | ok     |       4 |
|    4 | 22222222222 | 2017-01-10 | ok     |       7 |
+------+-------------+------------+--------+---------+
2 rows in set (0.00 sec)