Question

以下是我的js验证码以及表单提交事件。

问题在于，当我提交表单时，它会验证用户ID和名称，并且不会继续。但是，如果我删除登录ID验证，那么它只检查用户名并正确提交表单。

Firebug显示其验证登录ID并停止前进。如果登录ID可用，我从数据库得到“0”。

$(document).ready(function () {

    var value = $('#button input').val();
    var name = $('#button input').attr('name');

    $('#button input').remove();
    $('#button').html('<a href="#" class="cssSubmitButton" rel=' + name + '>' + value + '</a>');

    //global vars
    var form = $("#customForm");
    var name = $("#name");
    var nameInfo = $("#nameInfo");
    var login_id = $("#login_id");
    var login_idInfo = $("#login_idInfo");

    //On blur
    name.blur(validateName);
    login_id.blur(validatelogin_id);

    //On key press
    name.keyup(validateName);
    login_id.keyup(validatelogin_id);

    //On Submitting
    $('#button a').on('click', function () {

        var link = $(this);
        if (validateName() & validatelogin_id()) //it validates login id here and gets stuck; if i remove validatelogin_id() it processes the form perfectly
        {
            var str = $("form").serializeArray();

            $.ajax({
                url: 'load.php',
                data: str,
                type: 'POST',
                cache: 'false',
                beforeSend: function () {
                    link.addClass('loading');
                },
                success: function (data) {
                    link.removeClass('loading');
                    $('#button').css('display', 'none');
                    $('#success').css('display', 'block');

                },
                error: function (x, e) {

                }
            });

            return true
        } else {
            return false;
        }
    });

    //validation functions
    function validatelogin_id() {
        //if it's NOT valid
        if (login_id.val().length < 4) {
            $('#login_id_correct').css('display', 'none');
            login_id.addClass("error");
            login_idInfo.text("We want names with more than 3 letters!");
            login_idInfo.addClass("error");
            return false;
        }
        //if it's valid
        else {

            $.ajax({
                type: "post",
                url: "./include/check_user_name.php",
                data: "username=" + login_id.val(),
                success: function (data) {
                    if (data == 0) {
                        $('#login_id_correct').css('display', 'block');
                        login_id.removeClass("error");
                        login_idInfo.text("What's your name?");
                        login_idInfo.removeClass("error");
                        return true;
                    } else {
                        $('#login_id_correct').css('display', 'none');
                        login_id.addClass("error");
                        login_idInfo.text("Username already used!");
                        login_idInfo.addClass("error");
                        return false;

                    }
                }
            });
        }
    }

    function validateName() {
        //if it's NOT valid
        if (name.val().length < 4) {
            name.addClass("error");
            nameInfo.text("We want names with more than 3 letters!");
            nameInfo.addClass("error");
            return false;
        }
        //if it's valid
        else {
            name.removeClass("error");
            nameInfo.text("What's your name?");
            nameInfo.removeClass("error");
            return true;
        }
    }
});

Answer 1

假设您在onblur事件上调用validatelogin_id()

，此解决方案有效

valid_id = false; //declare a global flag

function validatelogin_id() {
        if (login_id.val().length < 4) {
            $('#login_id_correct').css('display', 'none');
            login_id.addClass("error");
            login_idInfo.text("We want names with more than 3 letters!");
            login_idInfo.addClass("error");
            valid_id = false; //set value of global flag           
        }
        else {

            $.ajax({
                type: "post",
                url: "./include/check_user_name.php",
                data: "username=" + login_id.val(),
                success: function (data) {
                    if (data == 0) {
                        $('#login_id_correct').css('display', 'block');
                        login_id.removeClass("error");
                        login_idInfo.text("What's your name?");
                        login_idInfo.removeClass("error");
                        valid_id = true; //set value of global flag
                    } else {
                        $('#login_id_correct').css('display', 'none');
                        login_id.addClass("error");
                        login_idInfo.text("Username already used!");
                        login_idInfo.addClass("error");
                        valid_id = false; //set value of global flag
                    }
                }
            });
        }
    }

然后更改此行

if (validateName() & validatelogin_id())

到

if (validateName() & valid_id)

AJAX在验证后未提交表单

1 个答案: