Question

我有一个这样的数组：

$post = array(    
    "name" => "John",
    "user" => "1" ,
    "title" => "hello" ,
    "uploader_0_name" => "pic.jpg",
    "uploader_0_status" => "done",
    "uploader_1_name" => "aaaa.jpg",
    "uploader_1_status" => "done",
    "uploader_2_name" => "Tulips.jpg",
    "uploader_2_status" => "failed",
    "uploader_count" => "3"
);

我想在另一个数组中使用uploader_[/d]_name和uploader_[/d]_name，例如：

    [0] => Array
    (
        [name] => pic.jpg
        [status] => done
    )
[1] => Array
    (
        [name] => aaaa.jpg
        [status] => done
    )
[2] => Array
    (
        [name] => Tulips.jpg
        [status] => failed
    )

在这种情况下，索引为0的数组应该有uploader_0_name,uploader_0_status

我在preg_match循环中尝试使用foreach进行了很多操作，但我无法成功

foreach ( $post as $key => $value ) {
$pattern = "/^uploader_[\d]_(name|status)$/";   
preg_match( $pattern , $key ,$matches[]);
}

P.S：不幸的是今天我看到了最好的答案和最好的方法被删除了，所以我添加了它，如果有任何人有这样的问题，可以使用：

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

    if (preg_match('/^uploader_(\d)_(name|status)$/', $key, $matches)) {
        $result[$matches[1]][$matches[2]] = $value;

    }
}

Answer 1

如果您不想使用正则表达式，请尝试此操作：

$newArr = array();

foreach($post as $key => $val) {
   $newKey = explode("_", $key);

   if (count($newKey) > 2) {
      //this is the status
      $innerValue = array_pop($newKey);

      //this is the numeric ID _2_ for example
      $innerKey   = array_pop($newKey);

      $newArr[$innerKey][$innerValue] = $val;
   }
}

Answer 2

有一种简单的方法可以不使用硬结构：

$post = array(
"name" => "John",
"user" => "1" ,
"title" => "hello" ,
"uploader_0_name" => "pic.jpg",
"uploader_0_status" => "done",
"uploader_1_name" => "aaaa.jpg",
"uploader_1_status" => "done",
"uploader_2_name" => "Tulips.jpg",
"uploader_2_status" => "failed",
"uploader_count" => "3"
);

//result array;
$arr = array();
//counter
$n = 0;

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

 if(strpos($key, '_name') != false){

    $arr[$n]['name'] = $value;

 }elseif(strpos($key, '_status') != false){

    $arr[$n]['status'] = $value;
    $n++;
 }

}

print_r($arr);

将数组键移动到另一个数组

2 个答案: