来自字符串的多维数组

Question

假设我有一个想要放入多维数组的字符串。

编辑：字符串中的子文件夹数量是动态的..从零个子文件夹到10个

<?php
       $string ="Folder1/Folder2/Folder3/filename1\n";
       $string .=" Folder1/Folder2/Folder3/filename2\n";
       $string .=" Folder4/Folder2/Folder3/filename3\n";
?>

我想要返回以下数组

<?php
 Array
(
    [Folder1] => Array
        (
            [Folder2] => Array
                (
                    [Folder3] => Array
                        (
                            [0] => filename1
                            [1] => filename2
                        )

                )

        )

    [Folder4] => Array
        (
            [Folder2] => Array
                (
                    [Folder3] => Array
                        (
                            [0] => filename3
                        )

                )

        )

)
?>

最有效的方法是什么？

为了好玩，假设这个数组将被发送到世界的另一端，它想要返回一个字符串。我们该怎么做呢？

Answer 1

您可以从 这个类（链接不再可用） 中借用一些代码，特别是 _processContentEntry 方法。

这是完成这项工作的方法的修改版本：

function stringToArray($path)
{
    $separator = '/';
    $pos = strpos($path, $separator);

    if ($pos === false) {
        return array($path);
    }

    $key = substr($path, 0, $pos);
    $path = substr($path, $pos + 1);

    $result = array(
        $key => stringToArray($path),
    );

    return $result;
}

输出

var_dump(stringToArray('a/b/c/d'));

会

array(1) {
  ["a"]=>
  array(1) {
    ["b"]=>
    array(1) {
      ["c"]=>
      array(1) {
        [0]=>
        string(1) "d"
      }
    }
  }
}

我想这就是你需要的:)

---

更新

根据您的评论，以下是处理由换行符分隔的字符串的方法：

$string = "Folder1/Folder2/Folder3/filename1\n";
$string .= " Folder1/Folder2/Folder3/filename2\n";
$string .= " Folder4/Folder2/Folder3/filename3\n";

// split string into lines
$lines = explode(PHP_EOL, $string);

// trim all entries
$lines = array_map('trim', $lines);

// remove all empty entries
$lines = array_filter($lines);

$output = array();

// process each path
foreach ($lines as $line) {
    // split each line by /
    $struct = stringToArray($line);

    // merge new path into the output array
    $output = array_merge_recursive($output, $struct);
}

print_r($output);

附： 要将此数组转换为字符串，只需调用json_encode，但我认为没有理由将其转换为数组然后再转换回原来的样子。

Answer 2

这可以通过另一种方式递归解决，方法是从数组的开头获取项目，当到达最后一个项目时返回它。

function make_tree( $arr ){
   if( count($arr) === 1){
      return array_pop( $arr );
   }else{
      $result[ array_shift( $arr )] = make_tree( $arr ) ; 
   }
   return $result;
}

$string  = "Folder1/Folder2/Folder3/filename1\n";
$string .= "Folder1/Folder2/Folder3/filename2\n";
$string .= "Folder4/Folder2/Folder3/filename3\n";

$string = trim( $string );

$files_paths = explode( PHP_EOL, $string);

$result = [];

foreach ($files_paths as $key => $value) {
   $parted = explode( '/', $value );
   $tree = make_tree( $parted );
   $result = array_merge_recursive( $result, $tree );
   
}
var_dump( $result );

Answer 3

我认为这是你想要的，

$string ="Folder1/Folder2/Folder3/filename1\n";
$string .="Folder1/Folder2/Folder3/filename2\n";
$string .="Folder4/Folder2/Folder3/filename3\n";


$string_array_1 = explode("\n", $string);

$array_need = array();

foreach($string_array_1 as $array_values)
{
        if($array_values)
        {
            $folders =  explode("/", $array_values);
            $array_need[$folders[0]][$folders[1]][$folders[2]][] = $folders[3];
        }
    }

print_r($array_need);