我的返回函数不适用于我的jQuery ajax

Question

I have this:

$("#upvote").click(function(){
    var up = parseInt(document.getElementById('voteScore').innerHTML);
    up++;
    document.getElementById('voteScore').innerHTML = up;
    $.ajax({
            type: 'POST',
        url: 'include/mysql_lib.php',
        data: {'data[]':['upvote','<?php echo $id; ?>', '<?php echo $uid; ?>']},
        dataType: "text",
        success: function(dataType) {
                if (dataType == "false") {
                    var up = parseInt(document.getElementById('voteScore').innerHTML);
                    up--;
                    document.getElementById('voteScore').innerHTML = up;
                }
        }
    });
});

The mysql_lib.php file (if an error is found) has a line like this:

return "false";

What am I doing wrong? I've never used jQuery before.

Answer 1

The AJAX function's server response (the 'dataType' variable in your code) stores whatever your PHP script wrote-out to the server. In your PHP script if you return "false" that will return a string from a function, but if you want to pickup the value in your JavaScript you should use echo "false" so that the JavaScript's response from the server will be false .

function test() {
    //do some work
    return "false";
}
echo test();//this will output "false" to the browser

function test() {
    echo "false";//this will output "false" to the browser
}

When running into issues like this it is a good idea to put a console.log(dataType) or alert(dataType) inside your AJAX callback function to see what is being output by your PHP. The response from your PHP script can also be viewed in most developer tools (like FireBug).

And a suggestion for ya. If you want to get into outputting more complex information from your PHP script, take a look at the PHP json_encode() function which makes communicating between PHP and JavaScript painless.