jfinal ajax url 接受参数

Question

String openid = WeiXinUtil.getOpenId(code, WeiXinUtil.APP_ID, WeiXinUtil.APP_SECRET);
if(!StringUtil.isBlank(openid)){
LotteryInfo hasLoty = LotteryInfo.dao.ltyInfo(openid, "1");
if(hasLoty!=null){
String states = hasLoty.getStr("states");
if(!StringUtil.isBlank(states)){
if(states.equals("0")){
//未兑奖的，进行兑奖...
//hasLoty.set("states", "1").update();
setAttr("msg", "兑奖成功");
setAttr("descp", hasLoty.getStr("lty_name"));
setAttr("ltyid",hasLoty.getInt("lty_id"));

render("/register/success.jsp");

这个是java方法

var phoneCode = '';
var ltyid= '${ltyid}';
$(function(){
$("#sure").click(function(){
$.ajax({
url:"/exh/getLoty/"+ltyid,
success:function(data){
if(data.result)
{
$(".state").text("已领奖");
$("#sure").hide();
}
}
});
});
这是ajax

public void getLoty() {
String ltyid = getPara(0,"");
if ("".equals(ltyid))
return;
new LotteryInfo().set("lty_id", ltyid).set("states", "1").update();
setAttr("result", "success");
renderJson();
}
点击的方法

ltyid取不到还是干什么

debug

求大神帮帮忙    怎么取到ltyid

Answer 1

Answer 2

我加了 type:"POST" 还是出现一样的情况