MySQL计数列值在其他列中的不同值

Question

我有一个表，我需要在其中返回两件事（最好是通过一个查询）：
1）每个日期的唯一ID数
2）行数，其中otherfield =“-”为唯一ID。 这意味着，如果同一天的id在otherfield输入值“-”的两倍，我希望它将其计为1。

示例表：

date | id | otherfield
----------
date1 | f  | abc
date1 | p  | -
date1 | p  | -
date2 | f  | abc
date2 | d  | dee

应该返回表：

date1 | 2 | 1
date2 | 2 | 0

目前我正在使用：

SELECT date, COUNT(DISTINCT `id`) AS id, SUM(IF(otherfield = '-', 1,0)) AS `undeclared` FROM mytable GROUP BY date

但这总结了otherfield的所有值，对id ='p'的条目进行了两次计数，我希望它仅对不同的id进行计数。

先感谢您。

Answer 1

只需使用一个count distinct的条件count distinct ：

select date, count(distinct `id`) as num_ids, 
       count(distinct case when otherfield = '-' then id end) as undeclared
from mytable 
group by date;

Answer 2

一种可能的方式：

select t1.date, t1.id_cnt, t2.dash_cnt from (
    select date, count( distinct id ) as id_cnt from your_table
    group by date
) t1
left join(
    select date, sum(dash_cnt) as dash_cnt from (
        select date, id, 1 as dash_cnt from your_table
        where otherfield = '-'
        group by date, id 
    ) t
    group by date
) t2 
on t1.date = t2.date

Answer 3

如何使用子查询？

SELECT 
    date, COUNT(DISTINCT id),
    (SELECT COUNT(*) FROM mytable WHERE otherfield = '-' WHERE id = t.id GROUP BY id)
FROM mytable t
GROUP BY date