php数组的总和（从mysql结果创建），具体取决于另一个mysql列中的mysql值

Question

一个名为18_7_ChartOfAccounts表如下所示：

ID | AccountNumber
-------------
1  | 2310
2  | 2380
3  | 2610

另一个名为2_1_journal表如下所示：

ID | Amount | DebitAccount
--------------------------
1  | 26.03  | 2310
2  | 200.00 | 2310
3  | 3.63   | 2380
4  | 119.83 | 2380
5  | 33.86  | 2610
6  | 428.25 | 2610

目的是获得如下结果：

DebitAccount 2310 total is: 226.03
DebitAccount 2380 total is: 123.46
DebitAccount 2310 total is: 462.11

在此示例中，226.03总计为26.03 + 200.00

起初mysql代码

$query = "SELECT j.Amount, j.DebitAccount FROM 18_7_ChartOfAccounts AS c LEFT JOIN 2_1_journal AS j ON (c.AccountNumber = j.DebitAccount)";
$sql = $db->prepare($query);
$sql->execute();
$data = $sql->fetchAll(PDO::FETCH_ASSOC);

使用print_r($data); 得到像这样的长数组列表

[31] => Array
    (
        [Amount] => 26.03
        [DebitAccount] => 2310

[32] => Array
    (
        [Amount] => 200.00
        [DebitAccount] => 2310

如果在mysql查询中使用SUM(j.Amount)则仅获得一个总数（假设Column Amount总数）。

用

foreach($data as $result){
if(strlen($result['Amount']) > 0 ) {
echo "Amount ". $result['Amount']. "Account name ". $result['DebitAccount']. "<br>";
print_r (array_sum($result));
}
}

得到这样的东西

Amount 123.97Account name 2310
2433.97Amount 26.03Account name 2310
2336.03Amount 200.00Account name 2310

有任何想法如何获得必要的结果（以粗体显示）吗？

更新资料

将$ query更改为

$query = "SELECT SUM(j.Amount), j.DebitAccount FROM 18_7_ChartOfAccounts AS c LEFT JOIN 2_1_journal AS j ON (c.AccountNumber = j.DebitAccount) group by j.DebitAccount";

与print_r($data); 得到这样的数组

Array
(
[0] => Array
    (
        [SUM(j.Amount)] => 
        [DebitAccount] => 
    )

[1] => Array
    (
        [SUM(j.Amount)] => 110900.16
        [DebitAccount] => 2310
    )

[2] => Array
    (
        [SUM(j.Amount)] => 3660.86
        [DebitAccount] => 2380
    )

与数组似乎所有的作品。 现在，将foreach更改为echo "Amount ". $result['SUM(j.Amount)']. " Account name ". $result['DebitAccount']. "<br>"; echo "Amount ". $result['SUM(j.Amount)']. " Account name ". $result['DebitAccount']. "<br>";

得到

Amount 110900.16 Account name 2310
Amount 3660.86 Account name 2380
Amount 85247.40 Account name 2610

似乎还可以。 谢谢

Answer 1

您正在做错事。 您可以通过MySql语句本身获取总和。

将aggrgate函数sum和group by子句一起使用。

像这样，

SELECT DebitAccount,sum(Account) from  2_1_journal group by DebitAccount

您的完整代码：

$query = " SELECT DebitAccount,sum(Account) as Total from  2_1_journal group by DebitAccount";
$sql = $db->prepare($query);
$sql->execute();
$data = $sql->fetchAll(PDO::FETCH_ASSOC);
foreach($data as $result){
if(strlen($result['Total']) > 0 ) {
echo "DebitAccount ". $result['DebitAccount']. "Total is: ". $result['Total']. "<br>";
print_r (array_sum($result));
}
}

Answer 2

SELECT DebitAccount, SUM(Amount) 
FROM 2_1_journal 
GROUP BY DebitAccount

Answer 3

您必须在查询中使用GROUP BY

SELECT DebitAccount, SUM(Amount) AS Amount FROM 2_1_journal GROUP BY DebitAccount