Running commands from PHP exec shows different result than from shell - php

Question

The PHP exec command is not executing the same as the shell's interaction.
cd /var/www/myfolder
zip -r /var/www/myfolder/temp/newfile.zip ./*
generates just a zip of files in the temp directory. However (simplified version):
$zip_dir = '/var/www/myfolder';
$temp_dir = $zip_dir . '/temp/';
chdir($zip_dir);
exec('zip -r ' . $temp_dir . 'newfile.zip ./*', $return);
generates the same zip but with the full path's of var and www (which results in two copies of myfolder so my file is twice as large as needed). The $return however has the same output as the command line execution. Both state only 15 files directories/folders were zipped. There is no mention of var or www in the PHP output.

Answer 1

I believe the chdir() command will not have any bearing on how the commands in exec() are run. So this might fix it:
$zip_dir = '/var/www/myfolder';
$temp_dir = $zip_dir . '/temp/';
$cmd = 'cd ' . escapeshellarg($zip_dir) . ' && zip -r ' . escapeshellarg($temp_dir . 'newfile.zip') . ' ./*';
exec($cmd, $return);
Note we always escape variables being passed to the command line.
But why not just zip within PHP?
<?php
$zip_target = "/var/www/myfolder";
$zip_file = "/var/www/myfolder/temp/newfile.zip";
$zip_temp = tempnam(sys_get_temp_dir(), 'a458');
$zip_obj = new \ZipArchive();
if ($zip_obj->open($zip_temp, ZIPARCHIVE::OVERWRITE)) {
$zip_obj->addGlob("$zip_target/**/*.*");
}
$zip_obj->close();
rename($zip_temp, $zip_file);