jquery按回车提交数据的代码示例

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
 

$(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

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'; 
                    } 
                } 
            }); 
    }); 
});