Question

我有一张桌子，我需要返回两件事（最好有一个查询）：
1）每个日期的唯一ID数量
2）其他字段的行数=＆＃34; - ＆＃34;用于唯一ID。这意味着，如果同一天的ID输入值的两倍＆＃34; - ＆＃34;在otherfield中，我希望将其视为一个。

示例表：

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

应该返回表格：

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

但是这总结了otherfield的所有值，计算了id =＆＃39; p＆＃39;的条目。两次，我希望它只计算不同的id。

提前谢谢。

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

mysql count其他列中不同值的列值

3 个答案: