Question

我有一个包含3个字段的表（tbl）：id，date和status。表中的示例信息可能如下所示：

id date status
 1 8-20 open
 2 8-20 open
 3 8-20 closed
 4 8-21 closed
 5 8-22 open
 6 8-23 open
 7 8-24 closed

我想编写一个返回数据的查询，如下所示：

date open closed
8-20 2    1
8-21 0    1
8-22 1    0
8-23 1    0
8-24 0    1

我可以得到的最接近的是在另一个选择中包含两个联合选择，如下所示：

select date from (
select date, count(id) from tbl where status = 'open' group by date
union
select date, count(id) from tbl where status = 'closed' group by date)
group by date

我认为它会涉及一个完整的外部联接，如果它不是SQLite来解释其中一个状态有0个条目但我不知道如何构造两个左连接的日子获得理想的结果。

Answer 1

运行以下查询：

select date, sum(case when status ='open' then 1 else 0 end) as open from test group by date

sqlfiddle.com/#!2/f0c533/4

并为关闭做同样的事。

Answer 2

您可以使用相关子查询：

SELECT date,
       (SELECT COUNT(*)
        FROM tbl AS t2
        WHERE t2.date = tbl.date
         AND status = 'open'
       ) AS open,
       (SELECT COUNT(*)
        FROM tbl AS t2
        WHERE t2.date = tbl.date
         AND status = 'closed'
       ) AS closed
FROM tbl

或者，按日期分组，并计算组中具有所需状态的行数：

SELECT date,
       total(status = 'open') AS open,
       total(status = 'closed') AS closed
FROM tbl
GROUP BY date

复杂的SQLite加入

2 个答案: