Question

我有两个桌子。第一个表注册录取。第二个有出口，如下所示：

 Table 1: Admissions
+----------+---------------------+---------+
| entry_id | join_date           | name    |
+----------+---------------------+---------+
|       26 | 2017-01-01 00:00:00 | James   |
|       29 | 2017-01-01 00:00:00 | Jan     |
|       27 | 2017-01-01 00:00:00 | Chris   |
|       28 | 2017-01-01 00:00:00 | Mary    |
|       22 | 2017-01-02 00:00:00 | Anna    |
|       21 | 2017-01-02 00:00:00 | Andy    |
|       24 | 2017-01-02 00:00:00 | Bob     |
|       20 | 2017-01-04 00:00:00 | Alice   |
|       23 | 2017-01-04 00:00:00 | Chris   |
|       25 | 2017-01-04 00:00:00 | Happy   |
+----------+---------------------+---------+

Table 2: Exits
+----------+---------------------+----------+
| entry_id | exit_date           | name     |
+----------+---------------------+----------+
|      322 | 2017-01-01 00:00:00 | Kay      |
|      344 | 2017-01-01 00:00:00 | Agnes    |
|      920 | 2017-01-02 00:00:00 | Andre    |
|      728 | 2017-01-02 00:00:00 | Mark     |
|      583 | 2017-01-03 00:00:00 | Alsta    |
|      726 | 2017-01-03 00:00:00 | Bull     |
|      816 | 2017-01-03 00:00:00 | Jane     |
|      274 | 2017-01-04 00:00:00 | Jack     |
|      723 | 2017-01-04 00:00:00 | Anna     |
|      716 | 2017-01-04 00:00:00 | Bill     |
+----------+---------------------+----------+

我正在寻找一种解决方案，以了解按日期分组的入场人数，退出人数和余额。

我正在寻找这个>

+---------------------+--------+--------+-----------+
| date                | joins  | exist  | net       |
+---------------------+--------+--------+-----------+
| 2017-01-01 00:00:00 |      4 |      2 |         2 |
| 2017-01-02 00:00:00 |      3 |      2 |         1 |
| 2017-01-03 00:00:00 |      0 |      3 |        -3 |
| 2017-01-04 00:00:00 |      3 |      3 |         0 |
+---------------------+--------+--------+-----------+

注意：可能会有几天录取，但是没有注册退出，反之亦然。

Answer 1

您在这里：

SELECT
  d,
  SUM(CASE WHEN t = 'j' THEN 1 ELSE 0 END) as joins,
  SUM(CASE WHEN t = 'x' THEN 1 ELSE 0 END) as exits,
  SUM(CASE WHEN t = 'j' THEN 1 ELSE 0 END) - SUM(CASE WHEN t = 'x' THEN 1 ELSE 0 END) as net
FROM
  (SELECT join_date as d, 'j' as t FROM admissions) j
  UNION ALL
  (SELECT exit_date as d, 'x' as t FROM exits) x
GROUP BY d

我们使用UNION ALL连接数据，并记下其类型联接或退出，并带有一个简单的字符，稍后我们可以进行比较

我们将其按d分组，每行给出一个日期，然后对有条件地查看其'j' oin或e 'x' it的结果求和。如果该行是j，则将1添加到跟踪该天的联接总数的列，依此类推

这唯一不给您的是没有连接或退出的日子。（例如2018-12-25, 0, 0, 0，因为圣诞节关闭，那天没有人做任何事情。）但是您没有并不是说你想要那些。

如果您确实想要带有日期的行，并且退出为0，联接为0，净值为0，那么我们必须做一些附加的魔术，这会更让人头疼/使其难以理解（因此我忽略了）

Answer 2

将执行以下操作：

    select 
        CASE WHEN join_date is not null THEN join_date 
             WHEN exit_date is not null THEN exit_date END as date,
        entry.cnt as joins,
        exit.cnt as exits,
        (extry.cnt - exit.cnt) as net
    FROM
        (select join_date, COALESCE(count(*), 0) as cnt from Admissions group by join_date) entry 
    FULL OUTER JOIN
        (select exit_date, COALESCE(count(*), 0) as cnt from Exits group by exit_date) exit 
    ON 
        entry.join_date=exit.exit_date
    ;

Answer 3

我没有找到答案。这是我一个朋友的回答，下面是MySQL版本：

select aa.date, IFNULL(aa.joins, 0) joins, IFNULL(bb.exits,0) exits, (IFNULL(aa.joins,0) - IFNULL(bb.exits,0)) net
from
(
  select join_date date, count(name) joins
  from Admissions
  group by join_date
) aa
left join
(
 select exit_date date, count(name) exits
 from Exits
 group by exit_date
) bb on aa.date = bb.date

UNION

select bb.date, IFNULL(aa.joins, 0) joins, IFNULL(bb.exits,0) exits, (IFNULL(aa.joins,0) - IFNULL(bb.exits,0)) net
from
(
  select join_date date, count(name) joins
  from Admissions
  group by join_date
) aa
right join
(
 select exit_date date, count(name) exits
 from Exits
 group by exit_date
) bb on aa.date = bb.date order by date;

从具有组的多个表进行计数

3 个答案: