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';
                    }

                }
            });
    });

});