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When using PHP's zip_read() function to handle ZIP files, it’s quite common to encounter the error message “supplied argument is not a valid Zip directory resource.” This article provides a detailed analysis of the reasons behind this error and explains how to correctly use zip_read(), as well as how to avoid and fix this issue.
zip_read() is a function provided by PHP’s Zip extension, used to read entries (files or directories) from a ZIP archive. The typical usage process involves:
Using zip_open() to open a ZIP file, which returns a resource handle.
Using zip_read() to read the next entry from that resource.
Processing the entry, such as retrieving the filename, size, etc.
Example function usage:
$zip = zip_open('example.zip');
if (is_resource($zip)) {
while ($entry = zip_read($zip)) {
echo zip_entry_name($entry) . "\n";
}
zip_close($zip);
}
This error indicates that the argument passed to the zip_read() function is not a valid ZIP directory resource. This usually means:
zip_open() returned false because the ZIP file couldn’t be opened successfully.
The argument passed to zip_read() is not the actual resource returned by zip_open().
The resource has already been closed or corrupted.
In short, zip_read() requires a valid ZIP resource handle as its argument—otherwise, this error will occur.
If the file path passed to zip_open() is incorrect or the file does not exist, zip_open() will return false instead of a resource.
Solution: Ensure that the file path is correct and that the file exists and is readable.
Example:
$zipFile = 'path/to/file.zip';
if (!file_exists($zipFile)) {
die("File does not exist!");
}
$zip = zip_open($zipFile);
if (!is_resource($zip)) {
die("Failed to open ZIP file!");
}
This might happen due to a logic error in the code, where a non-resource value is passed to zip_read().
Solution: Always check whether the return value of zip_open() is a resource before calling zip_read().
if (is_resource($zip)) {
while ($entry = zip_read($zip)) {
// Process entry
}
} else {
echo "Failed to open ZIP file.";
}
If zip_close() is called before zip_read(), the resource is already closed, causing the error.
Solution: Ensure the function calls are in the correct order. Do not call zip_read() after the resource has been closed.
If the PHP Zip extension is not enabled, related functions may fail.
Solution: Check your PHP configuration to ensure the Zip extension is installed and enabled.
php -m | grep zip
If zip is not listed, enable it in php.ini or install the extension.
<?php
$zipFile = 'm66.net/files/test.zip'; // Make sure to replace with actual file path
<p>// Open the ZIP file<br>
$zip = zip_open($zipFile);</p>
<p>if (!is_resource($zip)) {<br>
die("Failed to open ZIP file or the file format is incorrect.");<br>
}</p>
<p>while (($entry = zip_read($zip)) !== false) {<br>
echo "Filename: " . zip_entry_name($entry) . "\n";<br>
}</p>
<p>zip_close($zip);<br>
?><br>
The error “supplied argument is not a valid Zip directory resource” usually occurs because the provided argument is not a valid ZIP resource. To avoid this issue, you should:
Ensure the ZIP file exists and the path is correct.
Verify that zip_open() returns a resource, not false.
Use the resource correctly and avoid closing it prematurely.
Make sure the PHP Zip extension is installed and enabled.
By carefully troubleshooting and addressing these issues, you can effectively prevent this error. We hope this article helps you resolve any confusion when using zip_read().
