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I finally solved it with the following method. If the masters have a better way of writing, please feel free to enlighten me

    function get_vip_brand_list($uid = UID)
    {
        // 第一個數(shù)組
        $brand_list = config('sales_brand');
        // 第二個數(shù)組，反轉(zhuǎn)鍵和值
        $node       = array_flip(get_auth_node($uid,'sales.brand'));
        // 比較兩個數(shù)組的鍵名，并返回交集
        $vip_node   = array_intersect_key($brand_list, $node);
        return $vip_node;
    }

You can use the functions of the array_diff series to operate. You can decide by yourself whether to use array_diff_key or assoc for the specific business.

According to the value 5 4 1 contained in array 2, find the simplest way to write the unset of the above array 1 which does not exist and the key is not 5 4 1. Ha
means I don’t understand

<?php
$keys1 = array_keys($array1); // 獲取數(shù)組1key列表
$diffKeys = array_diff($keys1,$array2);// 結(jié)算數(shù)組1和數(shù)組2 key差集
foreach($diffKeys as $key){
    unset($array1[$key]);
}

Create a new array to store the values ??you want to retain. Then loop through array two, and then use the array_push function to push the values ??to be retained in array one into the newly created array.

$arr1 = array(
    1 => "伊凡木門", 
    2 => "夢天木門",
    3 => "大自然地板",
    4 => "尚品宅配",
    5 => "德國都芳漆",
    6 => "左右沙發(fā)"
);
$arr2 = array(5, 4, 1);
$keys = array_keys($arr1);
$remove = array_diff($keys, $arr2);
foreach ($remove as $key) {
    unset($arr1[$key]);
}
var_dump($arr1);

First flip array 2, and then find the intersection. I think your solution is the correct one

foreach($arr2 as $value) {
    foreach($arr1 as $key => $val) {
        if($value == $key) {
            unset($arr1[$key]);
        }
    }
}
print_r($arr1);

You can learn about the array_slice function