Question

我一直试图总结多个数组的值，但没有成功。我觉得我的大脑受不了了，因此我需要一些帮助或提示。

这是我到目前为止所取得的最好成绩：

$bani= array();
$iduri=array();
$total=0;
foreach($qp as $luna) {

    $nsd_id=$luna['nsd_id'];  // different ids every foreach
    $nsd_amount=$luna['nsd_amount']; // different values every foreach

    $nsd_id=explode("@",$nsd_id); // this makes them arrays
   $nsd_amount=explode("@",$nsd_amount); // same , it makes them arrays

    $iduri=array_merge($iduri,$nsd_id); // i tried getting making an array of all the ids it finds
    for($x=0;$x<count($iduri);$x++){
            $bani[$iduri[$x]] += intval($nsd_amount[$iduri[$x]]);
            $total += intval($nsd_amount[$x]);
        }

}
$iduri=array_unique($iduri,SORT_REGULAR);
print_r($iduri);

我尝试过将找到的每个ID组成一个数组，然后根据其位置将其求和到该数组中。

它确实可以累加一些东西，但是我没有得到正确的值，还有很多未定义的偏移量。

我觉得有比这简单的方法了，我真的很感谢一些建议。

谢谢！ image of the columns in the DB i'm using

如果结果可能是以下内容的数组，我将不胜感激

Array ( 
[1] => 0 
[2] => 2500
[3] => 62000 
[4] => etc 
[5] => etc 
[12] => etc 
[14] => etc )

（以id-s为键，将值的总和作为对应id的值）

qp的var_export

Mysqli_result::__set_state(Array( 'Current_field' => NULL, 'Field_count' => NULL, 'Lengths' => NULL, 'Num_rows' => NULL, 'Type' => NULL, ))

foreach（部分）内$ value的var_export

Array ( 'Id' => '329', 'Month' => 'May', 'Year' => '2019', 'Empid' => '124', 'Addedby' => 'PMSCLT10002', 'Nsd_id' => '1@2@3@4@5@12@14@15', 'Nsd_amount' => '0@0@0@0@0@0@4000@0', )
Array ( 'Id' => '303', 'Month' => 'April', 'Year' => '2019', 'Empid' => '124', 'Addedby' => 'PMSCLT10002', 'Nsd_id' => '1@2@3@4@5@12@14@15', 'Nsd_amount' => '0@0@2000@0@0@500@0@0', )

..等等

Answer 1

这是您可以做的摘录，

$nsdIds = [];
$result = [];
foreach ($qp as $key => $value) {
    // exploding Nsd_ids
    $nsdTemp = explode("@", $value['Nsd_id']);
    // exploding nsd_amounts
    $nsdAmount = explode("@", $value['Nsd_amount']);
    // saving all new nsd ids and only keep unique nsd ids
    $nsdIds = array_unique(array_merge($nsdIds, $nsdTemp));
    // I am now combining keeping nsd id as key and nsd amount as its value
    $temp1 = array_combine($nsdTemp, $nsdAmount);

    foreach ($nsdIds as $value1) {
        // for latest php version
        // $temp1 contains keys of nsd ids
        //  value1 contains nsdids we are sure that it will be in nsdids as we are already merging those ids
        //  then fetching amounf for that nsdids
        //$result[$value1] = ($result[$value1] ?? 0) + $temp1[$value1];
        // for php version less than 7
        $result[$value1] = (!empty($result[$value1]) ? $result[$value1] : 0) +  
                            (!empty($temp1[$value1]) ? $temp1[$value1] : 0);
    }
}
print_r($result);die;

Demo for php 7和demo for < php 7。

Answer 2

只要同一行中的nsd_id和nsd_amount在@之间具有相同数量的值，那么它应该起作用。如果不是，那么您需要isset来检查$nsd_amount[$key]：

foreach($qp as $luna) {
    $nsd_id = explode("@", $luna['nsd_id']);
    $nsd_amount = explode("@", $luna['nsd_amount']);

    foreach($nsd_id as $key => $id) {
        if(isset($result[$id])) {
            $result[$id] += $nsd_amount[$key];
        } else {
            $result[$id]  = $nsd_amount[$key];
        }
    }   
}

用不同的键求和多个数组

2 个答案: