是否可以在不遍历数组的情况下知道元素是否存在于数组中？

Question

我有一个结构相同的元素数组。 这是一个例子：

<?php
    $arr =
        [
            [
                "el_name" => "abcd",
                "el_data" => "raw_dataè_15264df156g18df",
            ],
            [
                "el_name" => "efgh",
                "el_data" => "raw_data_sd25g",
            ],
            [
                "el_name" => "ijkl",
                "el_data" => "raw_data_dfdfgn48",
            ],
            [
                "el_name" => "mnop",
                "el_data" => "raw_data_²545",
            ],
        ];
    
    $new_el = [
        "el_name" => "efgh",
        "el_data" => "raw_data_sd25g",
    ];

我想添加$new_el，前提是它尚未出现在$arr 中。如果php有array_element_exists()函数会很方便;-)

为了这个问题，我尝试了这个：

$exists = false;
foreach($arr as $el) {
    $equal=true;
    foreach(array_keys($el) as $key) {
        if(isset($new_el[$key]) && $new_el[$key] === $el[$key]) {
            null;
            //equality
        } else {
            $equal = false;
        }
    }
    if($equal === true) {
        $exists = true;
        break;
    }
}
if($exists === true ) {
    echo '$new_el already exists in $arr' . PHP_EOL;
} else {
    echo '$new_el doesn\'t still exists in $arr' . PHP_EOL;
    $arr[] = $new_el;
}

您认为还有更高效、“更简单”的方法吗？...

Answer 1

您可能正在寻找in_array() 函数。

in_array($new_el, $arr);

返回true。

Answer 2

function array_element_exists($new_el,$arr){
    $found = array_filter($arr, function($arr) use ($new_el) {
        return $arr['el_name'] === $new_el['el_name'] && 
            $arr['el_data'] === $new_el['el_data'];
    });
    return $found;
}

  
$exsist = ["el_name" => "efgh","el_data" => "raw_data_sd25g"];
$notExsist = ["el_name" => "newName","el_data" => "newData",];

//will not be added since it is already exist
if (!array_element_exists($exsist,$arr)) {
    $arr[] = $exsist;
}

//will be added because it is new and not exist in `$arr`
if (!array_element_exists($notExsist,$arr)) {
    $arr[] = $notExsist;
}