mysql - MySQL重用变量输出NULL

Question

我正在使用以下 SQL。column1、column2 给出的输出如 122.5014 和 4.2857；但是最后一列给出输出NULL虽然我期待48.9266。

SQL 输入：

---

SELECT (

SELECT @h := ( SUM( (
TIME_TO_SEC( TIMEDIFF(  `stop` ,  `start` ) ) /3600 ) )
)
) AS  `HoursWorked` , (

SELECT @w := ( TIMESTAMPDIFF( 
DAY , MIN(  `start` ) , MAX(  `stop` ) ) /7 )
) AS  `InWeeks` , (

SELECT (
((@w *40) - ( @h ))
)
) AS  `DutyDistance` 
FROM  `work_table`

输出：

HoursWorked | InWeeks | DutyDistance
------------------------------------
122.5014    | 4.2857  | NULL

Answer 1

当使用聚合函数计算然后在表达式中使用时，用户变量的输出是不可预测的。因此，您的查询不起作用。

示例：

mysql> create table tbl_so_q23870035( i int );
Query OK, 0 rows affected (0.48 sec)

mysql> insert into tbl_so_q23870035 values( 2 ), ( 4 ), ( 9 ), ( 6 ), ( 15 );
Query OK, 5 rows affected (0.13 sec)
Records: 5  Duplicates: 0  Warnings: 0

mysql> select @s:=sum(i) as s, @s*2 as s2 from tbl_so_q23870035;
+------+------+
| s    | s2   |
+------+------+
|   36 | NULL |
+------+------+
1 row in set (0.00 sec)

mysql> select s, s*2 as s2
    ->   from ( 
    ->          select sum(i) as s from tbl_so_q23870035 
    ->   ) aggregated_data;
+------+------+
| s    | s2   |
+------+------+
|   36 |   72 |
+------+------+
1 row in set (0.00 sec)

因此，您必须先将查询更改为 calculate HoursWorked，然后InWeeks在内部查询中更改，然后再DutyDistance在外部查询中进行计算。不需要用户变量。

示例：

select `HoursWorked`, `InWeeks`, ( `InWeeks` * 40 - `HoursWorked` ) AS `DutyDistance` 
from (
    select  SUM( TIME_TO_SEC( TIMEDIFF(  `stop`,  `start` ) ) / 3600 ) AS `HoursWorked`
         ,  TIMESTAMPDIFF( DAY , MIN(  `start` ), MAX(  `stop` ) ) / 7 AS `InWeeks` 
    FROM  `work_table`
) aggregated_data