匹配值时组合2个数组

Question

所以我有从2个不同的数据库获取的这些数组，我想在['item_name']和['name']匹配时将它们组合在一个数组中，然后从Array2获取['id']

我尝试做in_array，但由于它是多维的，我无法得到我想要的正确输出，我也试过foreach但是我也不能得到正确的输出或者我做错了，我已经不多了想法我怎么能做我想要的输出。

示例Array1：

Array
(
    [0] => Array
        (
            [item_id] => 1
            [item_name] => Bag
            [Color] => Purple
        )

    [1] => Array
        (
            [item_id] => 2
            [item_name] => Pencil
            [Color] => Yellow
        )

    [2] => Array
        (
            [item_id] => 3
            [item_name] => Tumbler
            [Color] => Blue
        )

    [3] => Array
        (
            [item_id] => 4
            [item_name] => Shirt
            [Color] => Red
        )

)

示例Array2：

Array
(
    [0] => Array
        (
            [id] => 11
            [name] => Bag
        )

    [1] => Array
        (
            [id] => 22
            [name] => Pencil
        )

    [2] => Array
        (
            [id] => 33
            [name] => Tumbler
        )

    [3] => Array
        (
            [id] => 44
            [name] => Shirt
        )

    [4] => Array
        (
            [id] => 55
            [name] => Paper
        )

    [5] => Array
        (
            [id] => 66
            [name] => Chair
        )

    [6] => Array
        (
            [id] => 4
            [name] => Notebook
        )

)

所以我的预期输出是：

Array
(
    [0] => Array
        (
            [id] => 11
            [name] => Bag
            [Color] => Purple
        )

    [1] => Array
        (
            [id] => 22
            [name] => Pencil
            [Color] => Yellow
        )

    [2] => Array
        (
            [id] => 33
            [name] => Tumbler
            [Color] => Blue
        )

    [3] => Array
        (
            [id] => 44
            [name] => Shirt
            [Color] => Red
        )
)

Answer 1

您可以使用第二个数组中的名称和ID创建一个映射，然后循环以修改第一个数组。

请考虑以下代码：

$a1 = array("item_id" => 1, "item_name" => "Bag", "Color" => "Purple");
$a2 = array("item_id" => 2, "item_name" => "Pencil", "Color" => "Yellow");
$a3 = array("item_id" => 3, "item_name" => "Tumbler", "Color" => "Blue");
$b1 = array("item_id" => 11, "item_name" => "Bag");
$b2 = array("item_id" => 22, "item_name" => "Pencil");
$b3 = array("item_id" => 33, "item_name" => "Tumbler");
$arr1 = array($a1, $a2, $a3);
$arr2 = array($b1, $b2, $b3);

$map = []; // here keys are names and value are the id 
foreach($arr2 as $elem)
    $map[$elem["item_name"]] = $elem["item_id"];


$ans = [];
foreach ($arr1 as $elem) {
    if (array_key_exists($elem["item_name"], $map))  {
        $elem["item_id"] = $map[$elem["item_name"]];
        $ans[] = $elem;
    }
}

echo print_r($ans);

通过这种方式，如果尝试嵌套for循环，您将获得O(n)而不是O(n^2)的复杂性

Answer 2

尝试使用foreach循环。检查foreach循环中是否有任何索引值与另一个数组匹配，然后创建一个包含所有这些值的新数组。你可以在这里使用counter来获取循环内数组的值。

Answer 3

我一直试图简化我的代码，我想我得到了它，感谢anju和David Winder，如果有人对我这样做感兴趣，这是如何：

index = 0;
foreach ($array1 as $val){
    foreach ($array2 as $val2){
        if ($val['item_id'] == $val2['id']){
            $filtered[$index]['id'] = $val2['id'];
            $filtered[$index]['name'] = $val2['name'];
            $filtered[$index]['color'] = $val['color'];
            $index++;
        }
    }
}