如何在没有页面源和没有JSON的情况下进行AJAX回读响应？

Question

I use the following code to submit a form:

$.ajax({
    url: $('#myForm').attr('action'),
    url: 'index.php',
    type: 'post',
    /*dataType: 'html json', *///tell the server that you expect json 
    data: {
        "num"     : $('#myNum').val(), 
    },
    success: function(response) {
        alert(response);
        if ( check_validate() == 1) {
            alert(response);
        }
        else {}
    }
});

Then in the PHP index.php I have:

if (isset($_POST["num"])) {
    $num = $_POST["num"];
    if($num == 1) 
        echo "sure!"; 
}

The JS popups with the sure! and the whole index.php page source code. Is there a way to remove the source code and just have the response show without setting the dataType to JSON?

Answer 1

Simply stop script execution after output:

//must be before any other output, eg at top of file
if (isset($_POST["num"])) {
    $num = $_POST["num"];
    if($num == 1) 
        echo "sure!";
        die(); 
}