我有兩個數組，我想從第一個數組中僅通過鍵搜索我在第二個數組中擁有的數組，作為第三個數組，我想打印結果：$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');預期結果：array('bench-press-rod'=>'',"adidas-classic-backpack"=>'93549559913');
3 回答
寶慕林4294392
TA貢獻2021條經驗 獲得超8個贊
您可以在 PHP 中的一行中執行此操作，array_flip用于交換第二個數組的鍵和值，然后array_intersect_key在匹配的鍵上合并兩個數組：
$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)));
輸出：
Array
(
[bench-press-rod] =>
[adidas-classic-backpack] => 93549559913
)
一只斗牛犬
TA貢獻1784條經驗 獲得超2個贊
我能想到的最簡單的方法是遍歷第二個數組并將第一個數組中的匹配鍵添加到輸出中。如果該項目不存在，則它會放入Not found輸出...
$output = [];
foreach ( $colldata2 as $item ) {
$output[$item] = $colldata[$item] ?? 'Not found';
}
print_r($output);
給..
Array
(
[bench-press-rod] =>
[adidas-classic-backpack] => 93549559913
)
撒科打諢
TA貢獻1934條經驗 獲得超2個贊
檢查這個。
$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);
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