其实jquery按回车提交数据是很简单的事情，我们只要检查测到用户按了回车就直接绑定事件.click就实现了提交了，在按钮上我们绑定ajax提交表单事件即可。
核心代码
复制代码 代码如下:
$(document).ready(function(){
$(按下回车的控件).keydown(function(e){
var curkey = e.which;
if(curkey == 13){
$(#回车事件按钮控件).click();
return false;
}
});
});
是用js的ajax功能同时支持回车事件
复制代码 代码如下:
document.onkeydown = function (e) {
var theevent = window.event || e;
var code = theevent.keycode || theevent.which;
if (code == 13) {
$(#login_submit).click();
}
}
$(document).ready(function() {
    //登录提交
    $(#login_submit).bind('click',function(){
        var username=$(#username).val();
        var password=$(#password).val();
$.ajax({
                type : 'post',
                url : './login.php',
                data: {type :'ajax',username:username,password:password},
                success : function(msg){
                    //alert(msg);
                    if(msg==-1){
                        alert('用户名不能为空');
                        $(#username).focus();
                    }
                    if(msg==-2){
                        alert('用户名不存在');
                        $(#username).val();
                        $(#username).focus();
                    }
                    if(msg==-3){
                        alert('密码不能为空');
                        $(#password).focus();
                    }
                    if(msg==-6){
                        alert('密码输入不正确');
                        $(#password).focus();
                    }
                    if(msg==1){
                        //alert('登录成功');
                        window.location.href='./index.php';
                    }
}
            });
    });
});