在PHP中获取实际的文档根目录？

Question

我不确定该脚本的确切运行位置，因此它必须能够从任何地方访问此目录。

我试图通过从目录中获取图像文件名来创建图像列表，对其进行过滤，直到只有我想要的图像格式，然后使用<img>标签显示它们。

最初的进展非常顺利。 输出HTML被证明是一个问题。

虽然我可以使用$_SERVER["DOCUMENT_ROOT"] to work with the directories in PHP, it's problematic to output that value as part of the path in the tag's src`属性$_SERVER["DOCUMENT_ROOT"] to work with the directories in PHP, it's problematic to output that value as part of the path in the tag's 。

这是我当前的代码：

$unkwown_files = scandir($_SERVER['DOCUMENT_ROOT'] . "/path/images");
          foreach($unkwown_files as $file) {
            $exploded_filename = explode(".", $file);
            $file_type = array_pop($exploded_filename);
            $accepted_filetypes = [
              "png",
              "jpg",
              "gif",
              "jpeg"
            ];
            $picture_names = [];
            if (in_array($file_type, $accepted_filetypes)) {
              $picture_names[] = $file;
            }
          }
          foreach($picture_names as $picture) {
            $path_to_image =  $_SERVER['DOCUMENT_ROOT'] . "/nodes/images" . $picture;
            echo '<img src="' . $path_to_image . '" class="upload_thumbnail"/>';
          }

Answer 1

从一开始就通过扩展名过滤文件的方法略有不同。

$dir = $_SERVER['DOCUMENT_ROOT'] . "/path/images/";
/*
    glob() searches for files/paths that match the pattern in braces and returns the results 
    as an array. 

    The '*' is a wildcard character as you would expect.

    The flag 'GLOB_BRACE' expands the string so that it tries to match each
    ( From the manual: GLOB_BRACE - Expands {a,b,c} to match 'a', 'b', or 'c' )

*/
$col = glob( $dir . "*.{jpg,png,gif,jpeg}", GLOB_BRACE );

/*
    Iterate through the results, with each one being the filepath to the file found.

    As the glob() function searched for the required types we don't need to check 
    if they are in the allowed types array.

    Because you do not wish to display the fullpath to the image, a relative path is 
    preferred - thus we remove, from the path, the document root.
*/
foreach( $col as $index => $file ){
    $path_to_image = str_replace( $_SERVER['DOCUMENT_ROOT'], '', $file );
    echo '<img src="' . $path_to_image . '" class="upload_thumbnail"/>';
}

Answer 2

试试看

foreach($picture_names as $picture) {
    $path_to_image = $_SERVER['DOCUMENT_ROOT'] . "/nodes/images/" . $picture;
    echo '<img src="' . $path_to_image . '" class="upload_thumbnail"/>';
}