如何在php中另一个数组的每个索引中插入数组值？

Question

我有两个数组，因为我需要插入最后一个键的每个索引和另一个数组键的值，即php中的值。 我的示例数组如下。 我正在使用codeigniter框架。

第一个数组：

Array
(
    [0] => stdClass Object
        (
            [customer_name] => Cash
            [ordernumber] => 6452424
            [product_name] => Bacardi Rum
            [quantity] => 1
            [unit_price] => 25.00
            [inv_discount] => 0.00
            [salesman_id] => 25,27
        )

    [1] => stdClass Object
        (
            [customer_name] => Cash
            [ordernumber] => 6452424
            [product_name] => Baileys
            [quantity] => 1
            [unit_price] => 15.00
            [inv_discount] => 0.00
            [salesman_id] => 28,29
        )

)

第二个数组：

Array
(
    [0] => 140140,150150
    [1] => 151151,05180518
)

我需要o / p：

Array
    (
        [0] => stdClass Object
            (
                [customer_name] => Cash
                [ordernumber] => 6452424
                [product_name] => Bacardi Rum
                [quantity] => 1
                [unit_price] => 25.00
                [inv_discount] => 0.00
                [salesman_id] => 25,27
                [salesman] => 140140,150150
            )

        [1] => stdClass Object
            (
                [customer_name] => Cash
                [ordernumber] => 6452424
                [product_name] => Baileys
                [quantity] => 1
                [unit_price] => 15.00
                [inv_discount] => 0.00
                [salesman_id] => 28,29
                [salesman] => 151151,05180518
            )

    )

谁能帮我，提出一些解决方案。

Answer 1

在这种情况下，您只有一个对象数组（ stdClass type ）和另一个数组。 回答您的问题，您只需执行如下所示的代码。

foreach ($secondArray as $key => $value) {
    $firstArray[$key]->salesman = $value;
}

要么

foreach ($firstArray as $key => $object) {
    $object->salesman = $firstArray[$key];
}

Answer 2

foreach($arrA as $key=>$val){
   $arrA[$key]['salesman'] = $arrB[$key];
}

Answer 3

只要两个数组的大小相同， array_map将是合适的：

$resultArray = array_map(function ($rowA, $rowB) {
    $rowA->salesman = $rowB;
    return $rowA;
}, $firstArray, $secondArray);