Javascript 和 AJAX，仅在使用alert()时有效

我的 JavaScript 遇到问题。它的行为似乎很奇怪。这就是正在发生的事情。我有一个表单，在用户提交后，它调用一个函数（onsubmit 事件）来验证提交的数据，如果有问题或者用户名/电子邮件已经在数据库中（这部分使用 ajax），它将返回 false并使用 DOM 显示错误。这是下面的代码。它的奇怪之处在于，它仅在我使用空警报（''）消息时才有效，没有它，它就不起作用。谢谢您的帮助。

//////////////////////////////////////

function httpRequest() {
    var xmlhttp;

    if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
        xmlhttp=new XMLHttpRequest();
    } else if (window.ActiveXObject) {
        // code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
    } else {
        alert("Your browser does not support XMLHTTP!");
    }

    return xmlhttp;
}

function validateRegForm(reg) {

    var isValidForm = true;
    var warningIcon = "";//for later in case we want to use an icon next to warning msg

    with(reg) {


        var regFormDiv = document.getElementById("registration_form");

        //Check if dynamic div exist, if so, remove it
        if(document.getElementById('disp_errors') != null) {
            var dispErrorsDiv = document.getElementById('disp_errors');
            document.getElementById('reg_form').removeChild(dispErrorsDiv);
        }           

        //Dynamically create new 'div'
        var errorDiv = document.createElement('div');
        errorDiv.setAttribute('id','disp_errors');
        errorDiv.setAttribute('style','background-color:pink;border:1px solid red;color:red;padding:10px;');
        document.getElementById('reg_form').insertBefore(errorDiv,regFormDiv);




        var xmlhttp = httpRequest();
        var available = new Array();
        xmlhttp.onreadystatechange = function() {
            if(xmlhttp.readyState == 4)
            {   
                var response = xmlhttp.responseText;
                if(response != "") {

                    //Return values
                    var newValue = response.split("|");
                    available[0] = newValue[0]; 
                    available[1] = newValue[1]; 
                }
            }
        }

        xmlhttp.open("GET","profile_fetch_reg_info.php?do=available&un="+u_username.value+"&email="+u_email.value+"",true);
        xmlhttp.send(null);


        alert(' '); ////////////WITHOUT THIS, IT DOESN'T WORK. Why?

        if(available[1] == "taken") {
            errorDiv.innerHTML += warningIcon+'Username is already taken!<br />';
            isValidForm = false;
        } else if(u_username.value.length < 4){
            errorDiv.innerHTML += warningIcon+'Username must be more than 4 characters long!<br />';
            isValidForm = false;
        } else if(u_username.value.length > 35) {
            errorDiv.innerHTML += warningIcon+'Username must be less than 34 characters long!<br />';
            isValidForm = false;
        }


        if(available[0] == "taken") {
            errorDiv.innerHTML += warningIcon+'Email address entered is already in use!<br />';
            isValidForm = false;
        } else if(u_email.value == ""){
            errorDiv.innerHTML += warningIcon+'Email address is required!<br />';
            isValidForm = false;
        } else {
            //Determine if email entered is valid
            var filter = /^([a-zA-Z0-9_\.\-])+\@(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
            if (!filter.test(u_email.value)) {
                errorDiv.innerHTML += warningIcon+'Email entered is invalid!<br />';
                u_email.value = "";
                isValidForm = false;
            }
        }


        if(u_fname.value == ""){
            errorDiv.innerHTML = warningIcon+'Your first name is required!<br />';
            isValidForm = false;
        }

        if(u_lname.value == ""){
            errorDiv.innerHTML += warningIcon+'Your last name is required!<br />';
            isValidForm = false;
        }



        if(u_password.value.length < 4){
            errorDiv.innerHTML += warningIcon+'Password must be more than 4 characters long!<br />';
            isValidForm = false;
        } else if(u_password.value.length > 35) {
            errorDiv.innerHTML += warningIcon+'Password must be less than 34 characters long!<br />';
            isValidForm = false;
        } else if (u_password.value != u_password2.value) {
            errorDiv.innerHTML += warningIcon+'Password and re-typed password don\'t match!<br />';
            isValidForm = false;
        }


        if(u_squestion.value == ""){
            errorDiv.innerHTML += warningIcon+'A security question is required!<br />';
            isValidForm = false;
        }

        if(u_sanswer.value == ""){
            errorDiv.innerHTML += warningIcon+'A security answer is required!<br />';
            isValidForm = false;
        }

        if(u_address.value == ""){
            errorDiv.innerHTML += warningIcon+'Address is required!<br />';
            isValidForm = false;
        }

        if(u_city.value == ""){
            errorDiv.innerHTML += warningIcon+'City is required!<br />';
            isValidForm = false;
        }

        if(u_state.value == ""){
            errorDiv.innerHTML += warningIcon+'State is required!<br />';
            isValidForm = false;
        }

        if(u_zip.value == ""){
            errorDiv.innerHTML += warningIcon+'Zip code is required!<br />';
            isValidForm = false;
        }

        if(u_phone.value == ""){
            errorDiv.innerHTML += warningIcon+'Phone number is required!<br />';
            isValidForm = false;
        }

        if(isValidForm == false)
            window.scroll(0,0);

        return isValidForm;
    }

}

The alert()有帮助，因为这会延迟该函数中剩余 JavaScript 的处理（来自alert()向下到底部），留出足够的时间来完成 AJAX 请求。 AJAX 中的第一个字母代表“异步”，这意味着（默认情况下）响应将在“未来的某个时刻”出现，但不会立即出现。

一个修复（你应该not实现）是使处理同步（通过改变第三个参数open() to be false）这将停止进一步处理您的脚本（以及整个网页），直到请求返回。这很糟糕，因为它会有效地冻结 Web 浏览器，直到请求完成。

正确的解决方法是重构代码，以便任何依赖于 AJAX 请求结果的处理都进入onreadystatechange函数，并且不能位于发起AJAX请求的主函数中。

通常处理这种情况的方法是修改您的 DOM（在发送 AJAX 请求之前）以使表单只读并显示某种“正在处理”消息，然后在 AJAX 响应处理程序中，如果一切正常（服务器正确响应）and验证成功）调用submit()在表单上，​​否则使表单再次可写并显示任何验证错误。