Google visualization 无法通过Ajax调用加载google Visualization/graph？

是否有一种通过Ajax调用创建google图表的特殊方法，与静态方法不同

我生成的HTML是正确的，因为它将从普通HTML文件加载，但当我调用Ajax时，图形中的数据不会显示

---

我正在使用google.setOnLoadCallback和google.load'visualization'，1'，{packages:['table']}

您需要从ajax调用中获取数据，然后将其放入可视化函数中。 这是我的密码：

google.load('visualization', '1', { packages: ['corechart'] });
google.setOnLoadCallback(OnLoad);   

var url = '/Charting/GetData';
function OnLoad() {
    $.ajax({
        url: url,
        dataType: 'json',          
        success: function (response) {
            drawVisualization(response);
        }
    });
};

function drawVisualization(response) {
   var chart = new google.visualization.ColumnChart(
       document.getElementById('visualization'));

   var data = new google.visualization.DataTable(response);   
   chart.draw(data);
};

我还建议您使用此类生成正确的JSON响应：

public class ChartHelper
{
    public ColInfo[] cols { get; set; }
    public DataPointSet[] rows { get; set; }    
}

public class ColInfo
{
    public string id { get; set; }
    public string label { get; set; }
    public string type { get; set; }
}

public class DataPointSet
{
    public DataPoint[] c { get; set; }
}

public class DataPoint
{
    public object v { get; set; } // value
    public string f { get; set; } // format
}

然后你可以这样使用它：

[ActionName("data")]
public JsonResult Data()
{
   Random r = new Random();
   var graph = new ChartHelper
   {
       cols = new ColInfo[] {
           new ColInfo { id = "A", label = "Name", type = "string" },
           new ColInfo { id = "B", label = "Value", type = "number" },
       },
       rows = new DataPointSet[] {
           new DataPointSet { 
               c = new DataPoint[] 
               {
                   new DataPoint { v = "Name" },
                   new DataPoint { v = r.NextDouble()},
               }},
           new DataPointSet {
               c = new DataPoint[]
               {
                   new DataPoint { v = "Name2" },
                   new DataPoint { v = r.NextDouble()},
               }},
           new DataPointSet {
               c = new DataPoint[]
               {
                   new DataPoint { v = "Name3" },
                   new DataPoint { v = r.NextDouble()},
               }} 
            }
        };

        return Json(graph, JsonRequestBehavior.AllowGet);
    }