PHP:键相同的子数组的所有值之和

Question

注释:这不仅仅是对电话时间的求和，电话时间需要通过函数单独计算。

我目前有以下代码

$i = array(1,2,3,4,5,6,7,8,9,10,11,12);

$j = array();

foreach ($i as $item) {
    $month_query_phonetime = $conn->query("SELECT phone_time FROM bigscreen_daily_data WHERE MONTH(work_date) = {$item} AND YEAR(work_date) = YEAR(CURDATE()) AND user_id = 28");

    while ($dataitem = $month_query_phonetime->fetch_object()) {
        $j[][$item] = point_for_phone_time($dataitem->phone_time);
    }
}

print_r($j);

这给了我如下的结果

    Array
    (
        [0] => Array
            (
                [1] => 0
            )

        [1] => Array
            (
                [1] => 10
            )

        [2] => Array
            (
                [2] => 48
            )

        [3] => Array
            (
                [2] => 24
            )

        [4] => Array
            (
                [3] => 48
            )
        [4] => Array
            (
                [3] => 48
            )

    )

我试图将所有[1] => value、[2]=> value等的和作为一个变量，并将其传递给$j[][$item]中的工业项目，以便用于进一步的处理。

我试过

while ($dataitem = $month_query_phonetime->fetch_object()) {
    $m = $m + point_for_phone_time($dataitem->phone_time);
}

$j[][$item] = $m;

它给了我如下的价值

Array
(
    [0] => Array
        (
            [1] => {value1}
        )

    [1] => Array
        (
            [2] => {value1+value2}
        )
)

为了得到想要的结果，这是正确的格式来处理这个问题。

Answer 1

最后，将查询附加到我希望它出现的循环中，现在它将显示预期的结果。

foreach ($data as $obj)
{

    $month_query_phonetime = $conn->query("SELECT phone_time FROM bigscreen_daily_data WHERE MONTH(work_date) = {$obj->mon} AND YEAR(work_date) = YEAR(CURDATE()) AND user_id = {$userid} ");

    $m = 0;

    while ($dataitem = $month_query_phonetime->fetch_object()) {
        $m = $m + point_for_phone_time($dataitem->phone_time);
    }

    $obj->prog_point = $obj->prog_point + $m;
}

Answer 2

如果我正确地理解了并且您试图为每个phone_time求和每个user_id，为什么不让数据库为您做繁重的工作，并使用聚合查询：

SELECT   user_id, SUM(phone_time) AS sum_phone_time
FROM     bigscreen_daily_data
GROUP BY user_id

它的执行速度可能也比遍历user_id、查询每个phone_time并自己对它们进行求和更快。

Answer 3

让MySQL按月分组，并将数据集限制在今年：

SELECT   MONTH(work_date) AS m,SUM(phone_time) AS sum_phone_time
FROM     bigscreen_daily_data
WHERE    user_id = 28 AND work_date >= MAKEDATE(year(now()),1);
GROUP BY m