Array.prototype.find() 在 PHP 中 returns 引用

Question

我正在尝试在 javascript 中创建一个类似于 Array.prototype.find() 的函数，但对于 PHP.

我的数组有这样的结构：

$array = [
     ["id" => 54, "type" => 5, "content" => [
          ["id" => 99, "type" => 516],
          ["id" => 88, "type" => 464],
          ["id" => 41, "type" => 845]]
     ],
     ["id" => 54, "type" => 55, "content"=> [
          ["id" => 54, "type" => 578],
          ["id" => 54, "type" => 354],
          ["id" => 75, "type" => 458]]
     ],
     ["id" => 65, "type" => 59, "content" => [
          ["id" => 87, "type" => 5454],
          ["id" => 65, "type" => 245],
          ["id" => 65, "type" => 24525]]
     ]
];

然后我创建了我的函数来像 javascript 中那样进行搜索：

function array_find($array, $function){
    foreach($array as $value){
        if($function($value)){
            return $value;
        }
    }
}
$id=54;
$type=55;
$mycontent = array_find(
    $array,
    function($foo) {
        global $id;
        global $type;
        return $foo["id"] == $id && $foo["type"] == $type;
    }
)["content"];

它正在工作。但是我希望返回的结果是一个引用而不是一个值，所以我稍后可以在返回的数组中add/remove个元素。

我该怎么做？

Answer 1

我会尝试在不完全放弃您的原始代码的情况下回答这个问题。

如果你想return引用一个符合条件的行，那么我能想到的唯一方法就是

将您的数组数组转换为对象数组，
在函数名声明中添加&，并且
使用= &在全局范围内声明引用变量。

调整后的代码：(Demo)

function &getRowReference($array, $fn) {
    foreach ($array as $row) {
        if ($fn($row)) {
            return $row->content;
        }
    }
}

function qualifyingRow($row) {
    global $id;
    global $type;
    return $row->id == $id && $row->type == $type;
}

$id = 54;
$type = 55;

$array = [
    (object)[
        "id" => 54,
        "type" => 5,
        "content" => [
            ["id" => 99, "type" => 516],
            ["id" => 88, "type" => 464],
            ["id" => 41, "type" => 845]
        ]
    ],
    (object)[
        "id" => 54,
        "type" => 55,
        "content" => [
            ["id" => 54, "type" => 578],
            ["id" => 54, "type" => 354],
            ["id" => 75, "type" => 458]
        ]
    ],
    (object)[
        "id" => 65,
        "type" => 59,
        "content" => [
            ["id" => 87, "type" => 5454],
            ["id" => 65, "type" => 245],
            ["id" => 65, "type" => 24525]
        ]
    ]
];

$ref = &getRowReference($array, 'qualifyingRow');

$ref = 'this has been changed';

var_export($array);

我认为我自己不会费心去做所有这些努力。我可能会将数据结构保留在其原始状态，return 符合条件的行的索引，然后使用索引访问、改变和存储数据。

这是一个相关页面，提供一些关于 returning 参考的额外评论和警告：Return by reference in PHP

Array.prototype.find() 在 PHP 中 returns 引用

Array.prototype.find() in PHP that returns reference

php

arrays

reference

function