嗨大家好我想知道我是否可以通过我的隐藏输入值来提供Google Chart中的数据，我有一些单选按钮用于更改要存储在隐藏输入中的数据，我已经检查了这个和我的用于更改数据的代码似乎可以正常工作，但图表不会重新绘制。如何触发重绘功能。谢谢。
google.setOnLoadCallback(drawChart);
function drawChart() {
var sample_data = document.getElementById('sample_data').value;
var data2 = eval("["+sample_data+"]");
var data = new google.visualization.arrayToDataTable(data2);
var options = {
hAxis: {title: 'Age', minValue: 0,maxValue:105},
vAxis: {title:'Savings', minValue: 0,maxValue:2500000},
width: 960, height: 300,
colors: ['#4a82bd'],
legend:'none'
};
var chart = new google.visualization.AreaChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
drawChart();
function ViewModel() {
var self =this;
self.rangeSelect = ko.observable('0%');
self.dataGetter = ko.computed(function(){
var range = self.rangeSelect();
if(range=='0%'){
return "['Age','Savings'],"+
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}
if(range=='30%'){
return "['Age','Savings'],"+
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}
if(range=='40%'){
return "['Age','Savings'],"+
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}
if(range=='50%'){
return "['Age','Savings'],"+
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}
if(range=='60%'){
return "['Age','Savings'],"+
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;
}
if(range=='70%'){
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}
});
}
//Activate knockout.js
ko.applyBindings(new ViewModel());
这是我的小提琴[{3}}
答案 0 :(得分：1)
我能够通过jquery调用drawChart函数设置Timeout函数来修复我的问题所有细节都可以在我的jsfiddle上找到：）
答案 1 :(得分：0)
您需要为调用drawChart函数的单选按钮分配“onchange”事件处理程序。如果隐藏输入的值同时被更改，您应该将drawChart调用添加到该代码的末尾，而不是创建新的事件处理程序，以确保以正确的顺序发生事件。