js提交form表单

<form action="/Enterprise/member" id="sendinviteid" method="post">

</form>

function sendinvite(){

var sendinviteid = $('#sendinviteid');
sendinviteid.on('submit',function(ev){

ev.preventDefault();

if(1==2){
return false;
}

$.ajax({
url:"/xxx/xxx",
type:'POST',
data:sendinviteid.serialize(),
dataType:'json',
success:function(d){

if(d.status == 1){

alert("操作成功")

}else{

alert("数据异常!请重试!")
}

},error:function(){

alert("数据异常!请重试!")

}
});
});
}