从AJAX响应函数返回到父JS脚本

Question

无法获取AJAX响应，以将值传递给父Javascript脚本

下面的代码基于我在这里阅读的内容，但是显然，我缺少一些东西

    //AJAX CALL, START
        var response;
        var ajax = new XMLHttpRequest();
        var url = "hooks/__global_ajax_bespoke.php?address="+ address +"&club="+ club;
        var method = "GET";
        var asynchronous = true;
        ajax.open(method, url, asynchronous);
        //sending ajax request
        ajax.send();
        //receiving response from hooks/__global_ajax_bespoke.php
        ajax.onreadystatechange = 

        function(){return(
            function(){
                if(this.readyState == 4 && this.status == 200){
                    response = this.responseText;
                    //alert (response+' TEST');
                    return response;
                }
            }
        )}();

    //AJAX CALL, END
    //ACTING ON RETURNED RESPONSE, START
        alert('SEE RESPONSE '+response);
        if(response == "correct" || response == "incorrect"){
            return show_error('club', 'Select a "Club"  from within the same postcode as the address.');        
        }
        else{
            return show_error('club', 'ERROR FLAG FOR DEBUGGING PURPOSES.');        
        }           
    //ACTING ON RETURNED RESPONSE, END

好的地方-我知道我的回应是如预期的那样从php文件返回的 不好的地方-根据在该论坛和其他几个地方的了解，我期望通过返回anon函数，它将把其返回值传递给父javascript，但不会 任何帮助，请:) TIA