第2个回答 推荐于2017-06-22
用ajax,
var url = "要传递的目标地址";
//var data = "id=2&name=xiaodengzi";
var data = {};
data["id"] = 2;
data["name"] = "xiaodengzi";
var dataType = "json";
$.get(
url,
data,
function(response){
var name = response['num'];
$("#cx").html(name);
},
dataType
);
====================================
public function b(){
if($_GET['id'] == 2){
$number = M("number");
$find_number = $number->where(1)->find();
//print_r($find_number);//Array ( [num] => 31 )
$find_number['num'] += 1;
$where['id'] = 1;
$save['num'] = $find_number['num'];
$number->where($where)->save($save);
print json_encode($find_number);
}
}本回答被网友采纳