Question

所以我有这些数组是从2个不同的数据库中提取的，当 ['item_name'] 和 ['name'] 时，我想将它们组合成一个数组strong>被匹配，然后从 Array2

获得 ['id']

我尝试执行in_array，但是由于它是多维的，所以我无法获得所需的正确输出，我也尝试了foreach，但是我也无法获得正确的输出，或者我做错了，我我不知道该怎么做我想要的输出。

Array1示例：

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
        )

)

Array2示例：

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

尝试使用foreach循环。检查是否在foreach循环中是否有任何索引值与另一个数组匹配，然后使用所有这些值创建一个新数组。您可以在此处使用counter来获取循环内数组的值。

Answer 2

您可以使用第二个数组中的名称和ID创建一个映射，然后循环修改第一个数组。

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);

这样，如果尝试嵌套的for循环，则可以实现O(n)而不是O(n^2)的复杂性

Answer 3

感谢anju和David Winder，我尝试早些简化我的代码，我想我明白了，如果有人对我的代码感兴趣，这是这样的：

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++;
        }
    }
}

值匹配时组合2个数组

3 个答案: