Question

我正在建立一个查询，该查询旨在显示每月两个日期变量的出现次数。我能够组合两个独立的查询：我计算每月的出现次数和组数，但是我不知道如何将这两个查询联接在一起，因为它们来自同一张表，并且仍然只显示一列的一个月。

谢谢大家的帮助！

格式：YYYY-MM-DD

|---------------------|------------------|
|   onboard_date      | offboard_date    |
|---------------------|------------------|
|    2019/01/15       | -                |
|---------------------|------------------|
|    2019/01/25       |    2019/02/15    |
|---------------------|------------------|
|    2019/02/13       |    2019/02/20    | 
|---------------------|------------------|
|    2019/02/18       | -                |
|---------------------|------------------|
|    2019/03/09       | -                |
|---------------------|------------------|

我尝试过的工作：

SELECT DATE_TRUNC('month', onboard_date) AS onboard_month,
        COUNT(*) as onboards
FROM lukla.trn_users trn
WHERE trn.company_name = 'amaro'
GROUP BY DATE_TRUNC('month', onboard_date)
ORDER BY DATE_TRUNC('month', onboard_date)

和

SELECT DATE_TRUNC('month', offboard_date) AS onboard_month,
        COUNT(*) as onboards
FROM lukla.trn_users trn
WHERE trn.company_name = 'amaro' AND offboard_date IS NOT NULL 
GROUP BY DATE_TRUNC('month', offboard_date)
ORDER BY DATE_TRUNC('month', offboard_date)

我想要的结果：


|--------------|------------|------------|
|    month     | onboards   | offboards  |
|--------------|------------|------------|
|    01        |      2     |      0     |
|--------------|------------|------------|
|    02        |      2     |     2      |
|--------------|------------|------------|
|    03        |      1     |      0     |
|--------------|------------|------------|

Answer 1

横向联接使这一过程变得非常简单：

select date_trunc('month', v.dte) as month, sum(v.is_onboard) as onboards, sum(v.is_offboard) as offboards
from trn_users t cross join lateral
     (values (t.onboard_date, (t.onboard_date is not null)::int, 0),
             (t.offboard_date, 0, (t.offboard_date is not null)::int)
     ) v(dte, is_onboard, is_offboard)
where v.dte is not null
group by month
order by month;

Here是db <>小提琴。

Answer 2

您可以尝试完全连接两个派生表，一个派生表的数量，另一个派生表的数量。

SELECT coalesce(x.month, y.month) month,
       coalesce(x.count, 0) onboards,
       coalesce(y.count, 0) offboards
       (SELECT date_trunc('month', trn.onboard_date) month
               count(*) count
               FROM lukla.trn_users trn
               WHERE trn.company_name = 'amaro'
                     AND trn.onboard_date IS NOT NULL
               GROUP BY date_trunc('month', trn.onboard_date)) x
       FULL JOIN (SELECT date_trunc('month', trn.offboard_date) month
                         count(*) count
                         FROM lukla.trn_users trn
                         WHERE trn.company_name = 'amaro'
                               AND trn.offboard_date IS NOT NULL
                         GROUP BY date_trunc('month', trn.offboard_date)) y
                 ON y.month = x.month
       ORDER BY coalesce(x.month, y.month);

如何计算两个日期列的出现并将其仅显示在一个表中？

2 个答案: