Question

我试图在我的10个子文件夹中获取所有XML文件并解析它们。为此，我有以下代码：

public static function calculateEAXML ($dir) {
$dh  = opendir($dir);
$folderNames = array();
$arr = array();
while (false !== ($folderName = readdir($dh))) {
  if ( $folderName[0] == "." || (substr($folderName, -3) == "zip") ) {continue;}
  $folderNames[] = $folderName; 
  $dom = new DOMDocument;
  $dom->validateOnParse = true;
  foreach ($folderNames as $file) 
{
    if(is_dir($folderName)){ScanFiles::calculateEAXML($dir);}
    else{
    $df  = opendir($dir . $file);
    while (false !== ($file = readdir($df))) 
    {
        if ($file == "." || $file == ".."  ) {continue;}
        $dom->Load($dir . $folderName . "/" . $file);
        $arr[] = XML2Array::createArray($dom);
    }
  }
}
  return $arr;    
}
}

事情是它只在一个目录中解析文件而完全忽略另一个目录。有没有想法如何解析所有目录中的所有文件？

Answer 1

这是glob的用途：

foreach (glob("*.xml") as $filename) {
    // do sth with the file, $filename holds the current file
}

Answer 2

您的脚本存在许多问题，主要是foreach。

if(is_dir($folderName))：变量$folderName不存在。我认为你应该$file。
ScanFiles::calculateEAXML($dir);：您再次扫描同一目录，并且没有对响应做任何处理。
$df = opendir($dir . $file);：如果文件不是目录，则您尝试将其作为目录打开。
您的foreach循环位于while循环内。

这是尝试解决这些问题：

public static function calculateEAXML($dir) {
    $dh  = opendir($dir);
    $folderNames = array();
    $arr = array();

    // Get all files and folders in the current directory
    while (false !== ($folderName = readdir($dh))) {
        if ($folderName[0] == "." || (substr($folderName, -3) == "zip")) {
            continue;
        }
        $folderNames[] = $folderName;
    }

    foreach ($folderNames as $file) {
        $filename = $dir . '/' . $file;
        if(is_dir($filename)){
            $arr = array_merge($arr, ScanFiles::calculateEAXML($filename));
        } else {
            $dom = new DOMDocument;
            $dom->validateOnParse = true;
            $dom->Load($filename);
            $arr[] = XML2Array::createArray($dom);
        }
    }
    return $arr;    
}

PHP获取子目录中的所有文件

2 个答案: