从第一个数组返回键值，其键在第二个数组中

Question

我有两个数组，我想从第一个数组中仅通过键搜索我在第二个数组中拥有的数组，作为第三个数组，我想打印结果：$colldata=array("bench-press-rod"=>'',"adidas-classic-backpack"=>'93549559913',"adidas-classic-backpack-legend-ink-multicolour"=>'',"puma-suede-classic-regal"=>'93549920361,93549723753'); $colldata2=array(0 => 'bench-press-rod',1 => 'adidas-classic-backpack');预期结果：array('bench-press-rod'=>'',"adidas-classic-backpack"=>'93549559913');

Answer 1

您可以在 PHP 中的一行中执行此操作，array_flip用于交换第二个数组的键和值，然后array_intersect_key在匹配的键上合并两个数组：

$colldata=array("bench-press-rod"=>'',"adidas-classic-backpack"=>'93549559913',"adidas-classic-backpack-legend-ink-multicolour"=>'',"puma-suede-classic-regal"=>'93549920361,93549723753');

$colldata2=array(0 => 'bench-press-rod',1 => 'adidas-classic-backpack');

print_r(array_intersect_key($colldata, array_flip($colldata2)));

输出：

Array

(

    [bench-press-rod] => 

    [adidas-classic-backpack] => 93549559913

)

3v4l.org 上的演示

Answer 2

我能想到的最简单的方法是遍历第二个数组并将第一个数组中的匹配键添加到输出中。如果该项目不存在，则它会放入Not found输出...

$output = [];

foreach ( $colldata2 as $item ) {

    $output[$item] = $colldata[$item] ?? 'Not found';

}

print_r($output);

给..

Array

(

    [bench-press-rod] => 

    [adidas-classic-backpack] => 93549559913

)

Answer 3

检查这个。

$colldata=array("bench-press-rod"=>'',"adidas-classic-backpack"=>'93549559913',"adidas-classic-backpack-legend-ink-multicolour"=>'',"puma-suede-classic-regal"=>'93549920361,93549723753');

$colldata2=array(0 => 'bench-press-rod',1 => 'adidas-classic-backpack');

$result = [];

foreach ($colldata2 as $key => $value) {

    if (array_key_exists($value, $colldata)) {

        array_push($result,$colldata[$value]);

    }            

}

echo '<pre/>';

print_r($result);