Question

我想将movements的值加总，同时在同一个数组上保留day的唯一值。

基础数组：

array (607)
0 => array (2)
    day => "11" (2)
    movements => 1
1 => array (2)
    day => "11" (2)
    movements => 1
2 => array (2)
    day => "11" (2)
    movements => 1
3 => array (2)
    day => "12" (2)
    movements => 1
4 => array (2)
    day => "12" (2)
    movements => 1
5 => array (2)
    day => "12" (2)
    movements => 1

[...]

我期待的结果（对于每个day总和movements）：

array (607)
    0 => array (2)
        day => "11" (2)
        movements => 3
    1 => array (2)
        day => "12" (2)
        movements => 3    
    [...]

非常感谢任何帮助。

编辑：到目前为止我尝试了什么：

docs of QMainWindow：它还总结day，这不是我所期待的。

我自己的代码：

$arr1 = $stats['data']['vhc']; // contain ['day'] and ['movements']
$arr2 = $data['vhc']; // contain ['day'] and ['movements']
$sum = [];

if (!empty($arr1)) {
    $result = [];
    foreach ($arr1 as $key => $value) {
        if (isset($arr2[$key]['day']) && $arr2[$key]['day'] == $value['day']) {
            $result['day'] = $value['day'];
            $result['movements'] = $value['movements'] + $arr2[$key]['movements'];
        }
        array_push($sum, $result);
    }
}

Answer 1

您可以轻松使用foreach()和https://developers.google.com/analytics/devguides/collection/gtagjs/cross-domain

$final_array = [];

foreach($array as $arr){
  $final_array[$arr['day']]['day'] = $arr['day'];
  $final_array[$arr['day']]['movements'] = (isset($final_array[$arr['day']]['movements']))? $final_array[$arr['day']]['movements']+$arr['movements'] : $arr['movements'];    
}
$final_array = array_values($final_array);

print_r($final_array);

输出： - array_values()

Answer 2

此方法不会用相同的值覆盖结果数组数据。

代码：（Demo）

$array=[
    ['day'=>'11','movements'=>'1'],
    ['day'=>'11','movements'=>'1'],
    ['day'=>'11','movements'=>'1'],
    ['day'=>'12','movements'=>'1'],
    ['day'=>'12','movements'=>'1'],
    ['day'=>'12','movements'=>'1']
];
foreach($array as $row){  // iterate all rows
    if(!isset($result[$row['day']])){  // if first occurrence of day...
        $result[$row['day']]=$row;     // save the full row with day as the temporary key
    }else{                                    // if not the first occurrence of day...
        $result[$row['day']]['movements']+=$row['movements'];  // add movements value
    }
}
var_export(array_values($result));

输出：

array (
  0 => 
  array (
    'day' => '11',
    'movements' => 3,
  ),
  1 => 
  array (
    'day' => '12',
    'movements' => 3,
  ),
)

从多维数组中求和并合并重复值

2 个答案: