问题
I have compressed_file.zip on a site with this structure:
I want to extract all content from version_1.x folder to my root folder:
How can I do that? is possible without recursion?
回答1:
It's possible, but you have to read and write the file yourself using ZipArchive::getStream:
$source = 'version_1.x';
$target = '/path/to/target';
$zip = new ZipArchive;
$zip->open('myzip.zip');
for($i=0; $i<$zip->numFiles; $i++) {
$name = $zip->getNameIndex($i);
// Skip files not in $source
if (strpos($name, "{$source}/") !== 0) continue;
// Determine output filename (removing the $source prefix)
$file = $target.'/'.substr($name, strlen($source)+1);
// Create the directories if necessary
$dir = dirname($file);
if (!is_dir($dir)) mkdir($dir, 0777, true);
// Read from Zip and write to disk
$fpr = $zip->getStream($name);
$fpw = fopen($file, 'w');
while ($data = fread($fpr, 1024)) {
fwrite($fpw, $data);
}
fclose($fpr);
fclose($fpw);
}
回答2:
Look at the docs for extractTo. Example 1.
回答3:
I was getting similar errors to @quantme when using @netcoder's solution. I made a change to that solution and it works without any errors.
$source = 'version_1.x';
$target = '/path/to/target';
$zip = new ZipArchive;
if($zip->open('myzip.zip') === TRUE) {
for($i = 0; $i < $zip->numFiles; $i++) {
$name = $zip->getNameIndex($i);
// Skip files not in $source
if (strpos($name, "{$source}/") !== 0) continue;
// Determine output filename (removing the $source prefix)
$file = $target.'/'.substr($name, strlen($source)+1);
// Create the directories if necessary
$dir = dirname($file);
if (!is_dir($dir)) mkdir($dir, 0777, true);
// Read from Zip and write to disk
if($dir != $target) {
$fpr = $zip->getStream($name);
$fpw = fopen($file, 'w');
while ($data = fread($fpr, 1024)) {
fwrite($fpw, $data);
}
fclose($fpr);
fclose($fpw);
}
}
$zip->close();
}
来源:https://stackoverflow.com/questions/8102379/extract-folder-content-using-ziparchive