Question

我在PHP中看到了ZipArchive类，它允许您读取zip文件。但我想知道是否有办法迭代其内容而不先提取文件

Answer 1

作为对http://www.php.net/ziparchive的评论：

以下代码可用于获取所有文件名的列表一个zip文件。

<?php
$za = new ZipArchive(); 

$za->open('theZip.zip'); 

for( $i = 0; $i < $za->numFiles; $i++ ){ 
    $stat = $za->statIndex( $i ); 
    print_r( basename( $stat['name'] ) . PHP_EOL ); 
}
?>

Answer 2

http://www.php.net/manual/en/function.zip-entry-read.php

<?php
$zip = zip_open("test.zip");

if (is_resource($zip))
  {
  while ($zip_entry = zip_read($zip))
    {
    echo "<p>";
    echo "Name: " . zip_entry_name($zip_entry) . "<br />";

    if (zip_entry_open($zip, $zip_entry))
      {
      echo "File Contents:<br/>";
      $contents = zip_entry_read($zip_entry);
      echo "$contents<br />";
      zip_entry_close($zip_entry);
      }
    echo "</p>";
  }

zip_close($zip);
}
?>

Answer 3

我解决了这样的问题。

$zip = new \ZipArchive();

$zip->open(storage_path('app/'.$request->vrfile));

$name = '';

//looped through the zip files and got each index name of the files
//since I only wanted the first name which is the folder name I break the loop 
//after updating the variable $name with the index name and that's it

    for( $i = 0; $i < $zip->numFiles; $i++ ){
        $filename = $zip->getNameIndex($i);
        var_dump($filename);
        $name =     $filename;
        if ($i == 1){
            break;
        }
    }

    var_dump($name);

Answer 4

重复的问题。发布前搜索。 PHP library that can list contents of zip / rar files

    <?php

 $rar_file = rar_open('example.rar') or die("Can't open Rar archive");

 $entries = rar_list($rar_file);

 foreach ($entries as $entry) {
     echo 'Filename: ' . $entry->getName() . "\n";
     echo 'Packed size: ' . $entry->getPackedSize() . "\n";
     echo 'Unpacked size: ' . $entry->getUnpackedSize() . "\n";

     $entry->extract('/dir/extract/to/');
 }

 rar_close($rar_file);

 ?>

在PHP中，是否可以检查Zip文件的内容而不首先提取其内容？

4 个答案: