以字符串形式获取Array中对象的属性

Is there a one-line solution (specific function) you can use to get a specific property of all Objects in an Array as:

是否有一个单行解决方案（特定函数）可用于获取Array中所有对象的特定属性：

a (comma separated) String?
一个（逗号分隔）字符串？
the sum of all values?
所有价值的总和？

Example data (I know this isn't PHP code but it's an example of how the data looks like):

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);

Output I would like:

输出我想：

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

3 个解决方案

#1

no there is no build in function to get this done, but you can use the array_sum, array_map and the PHP implode Functions to get the work done.

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;

just edit the return $v['name']; line to your object specific return value. like return $v->name;

只需编辑返回$ v ['name'];行到您的对象特定返回值。比如返回$ v-> name;

//edit after comment

//评论后编辑

one liner solution

单线解决方案

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

#2

You want to loop through your users. Foreach is your friend.

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;

#3

Here is your solution, as you need.. Though, validation for integer value in Sum function is remained to implement.. you can add it manually..

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;
}

Edit: One Liner Solution

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))));
}

#1