What is the error message "readdir(): supplied argument is not a valid Directory resource" indicating in the PHP code provided?

The error message "readdir(): supplied argument is not a valid Directory resource" indicates that the argument passed to the readdir() function is not a valid directory handle. This typically occurs when the directory handle is not properly opened or initialized before calling the readdir() function. To solve this issue, make sure to open the directory using opendir() and store the directory handle in a variable before using it with readdir(). 

// Open the directory and store the handle in a variable
$dir_handle = opendir('/path/to/directory');

// Check if the directory handle is valid before using it with readdir()
if ($dir_handle) {
    // Read the directory contents using readdir()
    while (false !== ($file = readdir($dir_handle))) {
        echo $file . "<br>";
    }

    // Close the directory handle
    closedir($dir_handle);
} else {
    echo "Failed to open directory.";
}

Keywords

readdir Directory resource error message PHP code PHP functions

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