我如何选择var并更新ajax成功的值?

时间:2018-01-17 01:40:22

标签: javascript jquery

我想要选择一个变量,然后在我的ajax调用获得新的json时更新变量的值,但我无法弄清楚如何做到这一点。有人可以向我解释如何选择homePoints和awayPoints变量并将其替换为我的更新值吗?

var homePoints = game.total_points_bet_on_hometeam;
var awayPoints = game.total_points_bet_on_awayteam;

var myChart = new Chart(ctx, {
              type: 'doughnut',
              data: {
                labels: [homeTeam, awayTeam],
                datasets: [{
                  backgroundColor: [
                    hue,
                    hueTwo
                  ],
                  data: [homePoints, awayPoints]
                , borderWidth: 0
                }]
              },
              options: {
                    responsive: true
                ,   maintainAspectRatio: true
              }
            });

0 个答案:

没有答案