Question

我有一个像这样的dinamic值的数组。目标是合并数组，具有相同的代码和具有相同代码的值的总和。

$array = array(
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'15',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'15',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'10',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'5',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'15',
        ),
    ),
);

我想要这样的结果

$result = array(
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'30',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'35',
        ),
    ),
);

结果只有代码AA和BB，因为代码AC不存在于第一个索引上。然后将相同的值编码为总和。

由于

Answer 1

简单，这是一种方式

<?php
    $codes = array_column($array[0],"code");
    $out =array();
    foreach($array as $main)
    {
        foreach($main as $sub)
        {
            if(in_array($sub['code'], $codes))
            {
                if(isset($out[$sub['code']]))
                {
                    $out[$sub['code']]['cost']+=$sub['cost'];
                }else
                {
                    $out[$sub['code']] = $sub;
                }
            }       
        }       
    }
    print_r(array_values($out));
?>

测试结果

akshay@db-3325:/tmp$ cat test.php 
<?php
$array = array(
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'15',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'15',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'10',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'5',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'15',
        ),
    ),
);

    $codes = array_column($array[0],"code");
    $out =array();
    foreach($array as $main)
    {
        foreach($main as $sub)
        {
            if(in_array($sub['code'], $codes))
            {
                if(isset($out[$sub['code']]))
                {
                    $out[$sub['code']]['cost']+=$sub['cost'];
                }else
                {
                    $out[$sub['code']] = $sub;
                }
            }       
        }       
    }
    print_r(array_values($out));
?>

<强>输出

akshay@db-3325:/tmp$ php test.php 
Array
(
    [0] => Array
        (
            [code] => AA
            [name] => A Name
            [cost] => 30
        )

    [1] => Array
        (
            [code] => AB
            [name] => B Name
            [cost] => 35
        )

)

Answer 2

PHP code demo

<?php
ini_set("display_errors", 1);
$array = array(
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'15',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'15',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'10',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'5',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'15',
        ),
    ),
);
$newData=array();
array_map(function($arrayData) use(&$newData){
    foreach($arrayData as $data => $value)
    {
        $newData[]=$value;
    }
}, $array);

$result=array();
array_map(function($data) use(&$result){
    // array("AA","AB") in this array you can add your code for which you want to merge.
    if(in_array($data["code"], array("AA","AB")))
    {
        if(!isset($result[$data["code"]]))
        {
            $result[$data["code"]]=$data;
        }
        else
        {
            $result[$data["code"]]["cost"]+=$data["cost"];
        }
    }

}, $newData);
$result=array(array_values($result));
print_r($result);

<强>输出：

Array
(
    [0] => Array
        (
            [0] => Array
                (
                    [code] => AA
                    [name] => A Name
                    [cost] => 30
                )

            [1] => Array
                (
                    [code] => AB
                    [name] => B Name
                    [cost] => 35
                )

        )

)

Answer 3

<强>已更新

<?php

$collection = array(
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'15',
        ),
        array(
            'code'=>'AD',
            'name'=>'D Name',
            'cost'=>'45',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'15',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'10',
        ),
    ),
    array(
        array(
            'code'=>'AA',
            'name'=>'A Name',
            'cost'=>'5',
        ),
        array(
            'code'=>'AB',
            'name'=>'B Name',
            'cost'=>'10',
        ),
        array(
            'code'=>'AC',
            'name'=>'C Name',
            'cost'=>'15',
        ),
    ),
);

$newcollection=array();
$keysregistered=array();
$i=1;
foreach($collection as $list) {
    $collectionkeys = array();
    foreach($list as $arr) {
        $key=$arr['code'];
        $collectionkeys[]=$key;
        if(isset($keysregistered[$key])) {
             $oldtotal = $keysregistered[$key];
             $total = $oldtotal+$arr['cost'];
             $newcollection[$key]['cost'] = $total;
        } else {
             if($i==1) {
                 $newcollection[$key] = $arr;
                 $keysregistered[$key] = $arr['cost'];
             }
        }
    }

    $registeredkeys = array_keys($keysregistered);
    $diff = array_diff($registeredkeys,$collectionkeys);
    foreach($diff as $delkey) {
        unset($keysregistered[$delkey],$newcollection[$delkey]);
    }
$i++;
}

$finalarr = array_values($newcollection);

print_r($finalarr);


?>

现在代码已更新，它甚至会从第一个项目中删除AD，并为您提供唯一代码的总计。

Answer 4

使用array_column，array_map，call_user_func_array和array_merge函数的解决方案：

// $arr is your initial array

$merged = call_user_func_array("array_merge", $arr);

// getting `code` keys from the first nested item
$result = array_map(function($v){
    return [];
}, array_column($arr[0], 'cost', 'code'));

foreach ($merged as $item) {
    if (isset($result[$item['code']])) {
        if ($result[$item['code']]) {
            $result[$item['code']]['cost'] += $item['cost'];
        } else {
            $result[$item['code']] = $item;
        }
    }
}
$result = [array_values($result)];

print_r($result);

输出：

Array
(
    [0] => Array
        (
            [0] => Array
                (
                    [code] => AA
                    [name] => A Name
                    [cost] => 30
                )

            [1] => Array
                (
                    [code] => AB
                    [name] => B Name
                    [cost] => 35
                )
        )
)

合并多维数组然后求和值

5 个答案: