mysql计算其他列中不同值的列值

Question

我有一张表，我需要在其中返回两件事（最好使用一个查询）：
1) 每个日期的唯一 ID 计数
2) otherfield = "-" 为唯一 ID 的行数。这意味着，如果同一天的某个 id 在otherfield 中输入了两倍的值“-”，我希望它算作一。

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，计算 id='p' 的条目两次，我希望它只计算不同的 id。

提前谢谢你。

Answer 1

只需使用条件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

使用子查询怎么样？

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