Question

我有两个Z表。这些被命名为： MYSQL，attendance_summary。两个表都有相同的字段，但可能有不同的数据。

这是两张表中2016年8月的结果。

attendance_summary_cecb

使用这两个表，在一个查询中，我需要获取

mysql> SELECT emp_id, date, days, ot_days, ot_hours FROM attendance_summary;
+--------+------------+------+---------+----------+
| emp_id | date       | days | ot_days | ot_hours |
+--------+------------+------+---------+----------+
|      4 | 2016-08-01 |   10 |       0 |     20.0 |
|      4 | 2016-08-16 |   13 |       0 |     14.5 |
|     12 | 2016-08-01 |   12 |       0 |     10.0 |
|     12 | 2016-08-16 |   11 |       2 |     16.5 |
|     14 | 2016-08-01 |   12 |       2 |     15.0 |
|     14 | 2016-08-16 |   10 |       0 |     16.0 |
|     15 | 2016-08-01 |   11 |       0 |     16.0 |
|     15 | 2016-08-16 |    9 |       3 |     21.0 |
+--------+------------+------+---------+----------+
8 rows in set (0.00 sec)

mysql> SELECT emp_id, date, days, ot_days, ot_hours FROM attendance_summary_cecb;
+--------+------------+------+---------+----------+
| emp_id | date       | days | ot_days | ot_hours |
+--------+------------+------+---------+----------+
|      4 | 2016-08-01 |   15 |       0 |     20.0 |
|      4 | 2016-08-16 |   10 |       0 |     12.0 |
|     12 | 2016-08-01 |   12 |       1 |     10.0 |
|     12 | 2016-08-16 |   11 |       2 |     16.5 |
|     14 | 2016-08-01 |   12 |       2 |     15.0 |
|     14 | 2016-08-16 |   10 |       0 |     16.0 |
|     15 | 2016-08-01 |   10 |       0 |     16.0 |
|     15 | 2016-08-16 |    9 |       0 |     21.0 |
+--------+------------+------+---------+----------+
8 rows in set (0.00 sec)

这是我尝试的方式：

> From `attendance_summary` 
>  - Total Days
>  - Total OT Days
>  - Total OT Hours 
> 
> From `attendance_summary_cecb` 
>  - Total Days
>  - Total OT Days
>  - Total OT Hours

更新这是这两个表的表结构：

SELECT e.emp_id
     , e.full_name
     , MAX(days1) as days1
     , MAX(ot_days1) as ot_days1
     , MAX(ot_hours1) as ot_hours1
     , MAX(days2) as days2
     , MAX(ot_days2) as ot_days2
     , MAX(ot_hours2) as ot_hours2
     , SUM(days1+days2) AS olekma_days
     , SUM(ot_days1+ot_days2) AS olekma_ot_days
     , SUM(ot_hours1+ot_hours2) AS olekma_ot_hours       
     , SUM(days3+days4) AS cecb_days
     , SUM(ot_days3+ot_days4) AS cecb_ot_days
     , SUM(ot_hours3+ot_hours4) AS cecb_ot_hours        
FROM 
       (
        SELECT emp_id
           , CASE DAY(date) WHEN 1 THEN days ELSE 0 END as days1
           , CASE DAY(date) WHEN 1 THEN ot_days ELSE 0 END as ot_days1
           , CASE DAY(date) WHEN 1 THEN ot_hours ELSE 0 END as ot_hours1
           , CASE DAY(date) WHEN 16 THEN days ELSE 0 END as days2
           , CASE DAY(date) WHEN 16 THEN ot_days ELSE 0 END as ot_days2
           , CASE DAY(date) WHEN 16 THEN ot_hours ELSE 0 END as ot_hours2
        FROM attendance_summary
        WHERE MONTH(date) = 08 AND YEAR(date) = 2016

        UNION

        SELECT emp_id
         , CASE DAY(date) WHEN 1 THEN days ELSE 0 END as days3
         , CASE DAY(date) WHEN 1 THEN ot_days ELSE 0 END as ot_days3
         , CASE DAY(date) WHEN 1 THEN ot_hours ELSE 0 END as ot_hours3
         , CASE DAY(date) WHEN 16 THEN days ELSE 0 END as days4
         , CASE DAY(date) WHEN 16 THEN ot_days ELSE 0 END as ot_days4
         , CASE DAY(date) WHEN 16 THEN ot_hours ELSE 0 END as ot_hours4
        FROM attendance_summary_cecb
        WHERE MONTH(date) = 08 AND YEAR(date) = 2016
       ) as s
LEFT JOIN employees e USING (emp_id)
INNER JOIN categories c ON c.id = e.category_id AND c.salary_type = 'D'
WHERE e.category_id = 6
GROUP BY e.emp_id\G

但它对我不起作用。希望有人可以帮助我。

Answer 1

你的问题的答案很简单。顺便说一下，我不得不从底部创建2个缺失的表来确认你的错误。

当你做

时

select col1, col2 from t1
union
select col9 as fred, col11 as john from t2

或该主题与scads表的任何变体，唯一返回的列名来自联盟的第一个成员。

换句话说，弗雷德和约翰没有出来。

因此，派生表s中没有您的days3和days4。

然后你需要重新考虑它，但这回答了问题：p

来自名为UNION Syntax的MySQL手册页：

第一个SELECT语句中的列名用作返回结果的列名称。列出的选定列每个SELECT语句的相应位置应该相同数据类型。（例如，第一列选择的第一列语句应与选择的第一列具有相同的类型其他陈述。）

我如何在MYSQL中创建这个SELECT查询？

1 个答案: