如何从 MySQL 中的两个表中获取数据并计算每个表？答案

【问题标题】：How to fetch data from two tables in MySQL and count each one?如何从 MySQL 中的两个表中获取数据并计算每个表？
【发布时间】：2016-11-13 18:13:57
【问题描述】：

我怎样才能在一个查询中写出来？

我有两张像这样的表，称为（迟到）：

id     name     S_id
1       A         6
2       A         6
3       B         5
4       C         8
5       A         6
6       A         6
7       C         8
8       C         8

另一个喜欢这个叫（缺席）：

id     name     S_id
1       A         6
2       A         6
3       A         6
4       A         6
5       A         6
6       A         6
7       B         5
8       c         8

我想要这个表格的结果：

其中 (count late) 计算迟到的次数，而 (count missing) 计算缺勤的时间。

name    Count late    Count absent
 A          4              6
 B          1              1
 C          3              1

我试过这样的：

这没用！

SELECT 
*
FROM
(SELECT name, COUNT(*) AS '# count absent' FROM absent GROUP BY s_id)  t1 
INNER JOIN
(SELECT name, COUNT(*) AS '# count Late' FROM late   GROUP BY s_id)  t2
ON t1.s_id = t2.s_id ;

【问题讨论】：

标签： mysql

【解决方案1】：

使用两个表的并集：lates 和absents...然后将lates 和absents 的数量相加。

试试这个：

SELECT 
    SUM(tardies) as 'total_lates', SUM(absences) as 'total_absences', name, s_id
FROM
    ((SELECT
        COUNT(*) as 'tardies',
        0 as 'absences',
        name,
        s_id
    FROM 
        lates
    GROUP BY
        s_id
    )
UNION
    (SELECT 
        0 as 'tardies',
        COUNT(*) as 'absences',
        name,
        s_id
     FROM
        absents
     GROUP BY
        s_id
    )
)
as maintable
GROUP by s_id
ORDER BY name

【讨论】：

不...每人只有一条记录...每条记录都有“迟到计数”和“缺席计数”...主表有一个分组依据...跨度>

【解决方案2】：

如果您需要加入 s_id，您必须在子选择中选择这些列

  SELECT 
  *
  FROM
  (SELECT s_id, name, COUNT(*) AS '# count absent' 
      FROM absent GROUP BY s_id)  t1 
  LEFT JOIN
  (SELECTs_id,  name, COUNT(*) AS '# count Late' 
      FROM late   GROUP BY s_id)  t2
  ON t1.s_id = t2.s_id ;

否则结果选择不能被连接，因为没有此列

【讨论】：

但这将消除那些要么只是缺席要么只是迟到的名字。
@TimBiegeleisen thaks correte update the left join .. 最终 OP 可以使用一个表来保存所有的名字，并为避免丢失名字而迟到和缺席加入表

【解决方案3】：

SELECT 
  (case when t1.name is not null then t1.name else t2.name end) as name,t1.absent,t2.late
  FROM
  (SELECT name, COUNT(*) AS 'absent' 
      FROM absent GROUP BY name)  t1 
  FULL JOIN
  (SELECT  name, COUNT(*) AS 'late' 
      FROM late   GROUP BY name)  t2
  ON t1.name = t2.name ;

试试这个。我没试过。希望它有效。

【讨论】：

I haven't tried ...如果您尝试过，您会发现MySQL中没有完全连接。
谢谢它的工作，但没有完全加入我只使用了加入。

【解决方案4】：

您需要在此处进行完整的外部联接，以确保保留迟到或缺勤的人，但不能同时保留这两者。嗯，MySQL 没有内置的完全外连接支持，但可以模拟：

SELECT t1.name,
       t2.late_cnt,
       t1.absent_cnt
FROM
(SELECT s_id, name, COUNT(*) AS absent_cnt
  FROM absent GROUP BY s_id, name) t1
LEFT JOIN
(SELECT s_id, name, COUNT(*) AS late_cnt
  FROM late GROUP BY s_id, name) t2
    ON t1.s_id = t2.s_id
UNION
SELECT t1.name,
       t2.late_cnt,
       t1.absent_cnt
FROM 
(SELECT s_id, name, COUNT(*) AS absent_cnt
 FROM absent GROUP BY s_id, name) t1
RIGHT JOIN
(SELECT s_id, name, COUNT(*) AS late_cnt
 FROM late GROUP BY s_id, name) t2
    ON t1.s_id = t2.s_id

【讨论】：

【解决方案5】：

请尝试以下。它对我有用。

select 
   coalesce( t1.name,t2.name) name,coalesce( t1.late,0) ,coalesce(t2.[absent],0) 
   from 
      (select name,s_id, count(*) as 'late' from late group by s_id,name ) t1 
   FULL OUTER JOIN
      (select name,s_id, count(*) as 'absent' from [absent] group by s_id,name ) 
   t2 on t1.s_id = t2.s_id 
   order by name

【讨论】：

同样，您的查询将过滤掉仅缺席但未迟到的人。
现在这将返回两个表记录！请让我知道你的看法。谢谢！！
MySQL 中没有完全外连接。但也许 OP 并不关心这种极端情况。
我必须感谢您，因为我从您的查询中学到了一些新东西。谢谢！！