如何使php请求报告失败

Question

I'm trying to scan the files in a directory using a php request. This is my code right now:

$.ajax({
    type: "POST",
    url: "scandir.php",

    data: { dirname: "./test" },
    success: function( data ) {
        console.log( data );
    }   
    error: function() {
        console.log( "error" );
    }
}); 

---

scandir.php:

<?php
    $dirname = $_POST[ "dirname" ];
    $files = scandir( $dirname ) or die("Unable to scan directory " . $dirname );
    foreach ( $files as $filename ) {
    echo $filename . "
";
    }
?>

Problem is, the request always returns successfully, even if the directory "dirname" doesn't exist (it returns the text of the php directory-not-found messages). How do I make it return with an error when the directory is not found?

Answer 1

if(!empty($_POST["dirname"])) {
    $dirname = $_POST["dirname"];

    if(!is_dir($dirname)) {
        header('HTTP/1.0 404 Not Found');
        exit;
    }
    foreach(scandir($dirname) as $filename) {
        echo $filename . "
";
    }
}