Question

我有一个代码来搜索目录中的文件。我正在使用代码var_dump($search->foundFiles)在网页中显示结果，但我无法弄清楚如何找到正确的代码以便正确显示，结果是网址。

var_dump($search->foundFiles)的结果是：

数组（[0] =＆gt; archivednews / 2016-01-08 22h30 Zika Virus归档news.html [1] =＆gt; archivednews / 2016-01-07 22h30 Zika Virus archived news.html）

但是我希望它显示在一个列表中，其中包含指向找到的文件的可点击链接：

<ul>
   <li><a href="archivednews/2016-01-05 22h30 Zika Virus archived news.html">2016-01-05 22h30 Zika Virus archived news.html</a></li>
   <li><a href="archivednews/2016-01-04 22h30 Zika Virus archived news.html">2016-01-04 22h30 Zika Virus archived news.html</a></li>
   <li><a href="archivednews/2016-01-08 22h30 Zika Virus archived news.html">2016-01-08 22h30 Zika Virus archived news.html</a></li>
</ul>

这是完整的代码：

class searchFileContents{
     var $dir_name = '';//The directory to search
     var $search_phrase = '';//The phrase to search in the file contents
     var $allowed_file_types = array('php','phps');//The file types that are searched
     var $foundFiles;//Files that contain the search phrase will be stored here
     var $myfiles;

  function search($directory, $search_phrase){
    $this->dir_name = $directory;
    $this->search_phrase = $search_phrase;

    $this->myfiles = $this->GetDirContents($this->dir_name);
    $this->foundFiles = array();

    if ( empty($this->search_phrase) ) die('Empty search phrase');
    if ( empty($this->dir_name) ) die('You must select a directory to search');

    foreach ( $this->myfiles as $f ){
        if ( in_array(array_pop(explode ( '.', $f )),  $this->allowed_file_types) ){
            $contents = file_get_contents($f);
            if ( strpos($contents, $this->search_phrase) !== false )
                $this->foundFiles [] = $f;
        }
    }
    return $this->foundFiles;
  }

  function GetDirContents($dir){
   if (!is_dir($dir)){die ("Function GetDirContents: Problem reading : $dir!");}
   if ($root=@opendir($dir)){
       while ($file=readdir($root)){
           if($file=="." || $file==".."){continue;}
           if(is_dir($dir."/".$file)){
               $files=array_merge($files,$this->GetDirContents($dir."/".$file));
           }else{
           $files[]=$dir."/".$file;
           }
       }
   }
   return $files;
  }
}

//Example :

$search = new searchFileContents;
$search->search('E:/htdocs/AccessClass', 'class');
var_dump($search->foundFiles);

Answer 1

var_dump仅在开发时使用。它只是调试的帮手。应用程序完成后，您无法将其输出给用户。

您可以像var_dump一样打印数组的每个值：

foreach($search->foundFiles as $ffiles)
    echo "<a href='$ffiles'>$ffiles</a><br>";

这会显示：

2016-01-05 22h30 Zika Virus archived news.html

2016-01-04 22h30 Zika Virus archived news.html

2016-01-08 22h30 Zika Virus archived news.html

但阵列中的网址不是真正的链接，所以你必须解决这个问题。

如何设置通过var_dump

1 个答案: