Mysql：JOIN 2表中，SUM之一不正确

Question

这是两个表的结构

表A

+----+-----+----+----------------------+--------------------+----------+
| id | ... |....|   time_start         | time_end           |  total   | 
+----+-----+----+----------------------+--------------------+----------+
  1               2015-12-06 10:00:00    2015-12-06 12:00:00     200
  2               2015-12-07 10:00:00    2015-12-07 12:00:00     300              

表B

+----+----------+------+------+------+------+
| id | idTableA | val1 | val2 | val3 | val4 |   
+----+----------+------+------+------+------+
  1        1       10     10     10     10
  2        1       10     10     10     10
  3        2       10     10     10     10

目标如下：给定一个time_start和一个time_end date，显示总计的总和（表A）和val1，val2，val3，val4的总和

范例 ：

• time_start = 2015-12-01 00:00:00

• time_end = 2015-12-30 23:59:59

预期结果 ：总计= 500，val（1-4）等于120

我已经尝试过了：

$myquery = "";
$myquery .= "SELECT SUM(tableA.total) AS myTotal,";
$myquery .= "SUM(tableB.val1) + SUM(tableB.val2) + SUM(tableB.val3) + SUM(tableB.val4) AS myValTotal ";
$myquery .= "FROM tableA INNER JOIN tableB ON tableA.id = tableB.idTableA ";
$myquery .= "WHERE tableA.time_start >='".$dateStart."' AND tableA.time_end <='".$dateEnd."'";

val（1-4）的SUM正确，但合计的SUM不正确。

Answer 1

您必须检查想要重叠的数据范围是部分还是全部。

确定两个日期范围是否重叠

最好的测试方法是首先直接在数据库上复制查询。

在WHERE检查数据范围

SET @dateStart= '2015-12-01 00:00:00';
SET @dateEnd = '2015-12-30 23:59:59';

SELECT myTotal, myValTotal 
FROM 
    (
      SELECT SUM(tableA.total) AS myTotal
      FROM tableA
      WHERE tableA.time_start >= @dateStart
        AND tableA.time_end <=  @dateEnd ;
    ) T1
CROSS JOIN 
    (
        SELECT SUM(tableB.val1 + tableB.val2 + tableB.val3 + tableB.val4) AS myValTotal 
        FROM tableA 
        INNER JOIN tableB 
                ON tableA.id = tableB.idTableA
        WHERE tableA.time_start >= @dateStart
          AND tableA.time_end <=  @dateEnd
   ) T2;

Answer 2

您不应该在行的乘法上进行汇总。 而是独立地聚合两个表，然后将它们联接起来，例如：

select * from
(
SELECT SUM(tableA.total) AS myTotal
FROM tableA 
WHERE tableA.time_start <= @dateEnd 
  AND tableA.time_end >=  @dateStart
) x join (
SELECT SUM(tableB.val1) + SUM(tableB.val2) + 
       SUM(tableB.val3) + SUM(tableB.val4) AS myValTotal 
FROM tableB join  tableA ON tableA.id = tableB.idTableA
WHERE tableA.time_start <= @dateEnd 
  AND tableA.time_end >=  @dateStart
) y;

Answer 3

作为起点，这似乎更容易阅读...

$myquery = 
 "
 SELECT SUM(a.total) myTotal
      , SUM(b.val1 + b.val2 + b.val3 + b.val4) myValTotal
   FROM tableA a
   JOIN tableB b
     ON b.idTableA = a.id 
  WHERE a.time_start >='$dateStart' 
    AND a.time_end   <='$dateEnd'
 ";

Answer 4

SELECT  
SUM(tableB.val1) + SUM(tableB.val2) + SUM(tableB.val3) + SUM(tableB.val4) AS myValTotal,
(SELECT SUM(total) from tableA  where   tableA.time_start >='2015-12-01 00:00:00' AND tableA.time_end <= '2015-12-30 23:59:59') as myTotal 
FROM tableA  INNER JOIN tableB ON tableA.id = tableB.idTableA 
WHERE tableA.time_start >='2015-12-01 00:00:00' AND tableA.time_end <= '2015-12-30 23:59:59'

Answer 5

在加入数据之前先汇总数据，这样您就不会误认为值是多重的。

select sum(a.total) as mytotal, sum(b.sumval) as myvaltotal
from tablea a
left join
(
  select idtablea, sum(val1+val2+val3+val4) as sumval
  from tableb
  group by idtablea
) b on b.idtablea = a.id
where a.time_start >= @start and a.time_end <= @end;

这与SELECT子句中的子查询相同。 它更简单，可以绕开Juan Carlos Oropeza在以下评论中描述的问题。

select 
  sum(total) as mytotal, 
  sum((
    select sum(val1+val2+val3+val4)
    from tableb
    where idtablea = tablea.id
  )) as sumvaltotal
from tablea
where time_start >= @start and time_end <= @end;

Answer 6

您可以声明int类型的变量并存储总和值，然后再次对存储的值求和以获得总值

          declare  @val1 int
          declare @val2 int
          declare @val3 int
          declare @val4 int
          declare @newval int

            select @val1= SUM(isnull(val1,0)) , @val2 =
            sum(isnull(val2,0)), @val3=sum(isnull(val3,0)),@val4 = 
            sum(isnull(val2,0)) from TableB

            select @newval = @val1 +@val2+@val3+@val4

@newval将包括val1与val4之和