Question

是否有单行解决方案（特定功能）可用于获取Objects中所有Array的特定属性：

a（逗号分隔）String？
所有值的总和？

示例数据（我知道这不是PHP代码，但它是数据的示例）：

$user1 = { id: 1, name: 'u1', score: 10 }
$user2 = { id: 2, name: 'u2', score: 20 }
$admins = array($user1, $user2);

输出我想：

magicalArrayFunction($admins, 'name', ', ');
//SHOULD OUTPUT: "u1, u2"
magicalSumArrayFunction($admins, 'score');
//SHOULD OUTPUT: 30

Answer 1

没有功能构建来完成这项工作，但您可以使用array_sum，array_map和PHP implode函数来完成工作。

function magicalArrayFunction($array)
{
    return implode(',', array_map(function($v){
        return $v['name'];
    }, $array));
}

$usersString = magicalArrayFunction($admins);
echo $usersString;

function magicalSumArrayFunction($array)
{
    return array_sum(array_map(function($v){
        return $v['score'];
    }, $array));
}

$arraySum = magicalSumArrayFunction($admins);
echo $arraySum;

只需将return $v['name'];行编辑为特定于对象的返回值即可。比如return $v->name;

//评论后编辑

一个班轮解决方案

$array = ...
$userString = implode(',', array_map(function($v){return $v['name'];}, $array));
$arraySum = array_sum(array_map(function($v){return $v['score'];}, $array));

Answer 2

您想要遍历您的用户。 Foreach是你的朋友。

$user1 = json_decode( '{"id":1,"name":"u1","score":10}');
$user2 = json_decode( '{"id":2,"name":"u2","score":20}');
$admins = array($user1, $user2);

$names = array();
$score = 0;
foreach( $admins as $user) {
    $names[] = $user->name;
    $score += $user->score;
}
echo implode( ",", $names) . "\n";
echo $score;

Answer 3

这是您的解决方案，因为您需要..虽然，Sum函数中的整数值验证仍然可以实现..您可以手动添加它..

$user1 = '{"id":1,"name":"u1","score":10}';
$user2 = '{"id":2,"name":"u2","score":20}';
$admins = array($user1, $user2);

echo magicalArrayFunction($admins, 'name', ', ');
//SHOULD OUTPUT: "u1, u2"
echo "<br>";
echo magicalSumArrayFunction($admins, 'score');
//SHOULD OUTPUT: 30

function magicalArrayFunction($array, $column, $sep = ', ') {
    foreach($array as $jsons){
        $new_Arr[] = json_decode($jsons, true);
    }
    return implode($sep,array_column_custom($new_Arr, $column));
} 
function magicalSumArrayFunction($array, $column){
    foreach($array as $jsons){
        $new_Arr[] = json_decode($jsons, true);
    }
    return array_sum(array_column_custom($new_Arr, $column));
}

function array_column_custom($array, $column){
    foreach($array as $keys=>$inner_array){
        foreach($inner_array as $inner_keys=> $values){
            if($inner_keys == $column){
                $newArr[] = $values;
            }
        }
    }
    return $newArr;
}

编辑：一个班轮解决方案

$user1 = '{"id":1,"name":"u1","score":10}';
$user2 = '{"id":2,"name":"u2","score":20}';
$admins = array($user1, $user2);

// Fixed column name used here i.e. 'name','score'.. 
$userNames = implode(',', array_map(function($u){$u=json_decode($u, true); return $u['name'];}, $admins));
$totalScore = array_sum(array_map(function($u){ $u=json_decode($u, true); return $u['score'];}, $admins));

// If you need to call function..
echo magicalArrayFunction($admins, 'name', ', ');
echo magicalSumArrayFunction($admins, 'score');
function magicalArrayFunction($array, $column, $sep = ', ') {
    return implode($sep, array_map(function($u,$c){ $u=json_decode($u, true); return $u[$c];}, $array, str_split(str_repeat($column, count($array)),strlen($column)) ));    
} 
function magicalSumArrayFunction($array, $column){
    return array_sum(array_map(function($u,$c){ $u=json_decode($u, true); return $u[$c];}, $array,str_split(str_repeat($column, count($array)),strlen($column))));
}

以字符串形式获取Array中对象的属性

3 个答案: