Jquery 将带有google图表的外部页面加载到div中
以下是google chart in.php的代码:Jquery 将带有google图表的外部页面加载到div中,jquery,html,charts,external,Jquery,Html,Charts,External,以下是google chart in.php的代码: <html><head> <script type="text/javascript" src="https://www.google.com/jsapi"></script> <script type="text/javascript"> google.load('visualization', '1.0', {'packages':['co
<html><head>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1.0', {'packages':['corechart']});
google.setOnLoadCallback(drawChart);
function drawChart() {
var data = new google.visualization.DataTable();
data.addColumn('string', 'Topping');
data.addColumn('number', 'Slices');
data.addRows([ ['Mushrooms', 3], ['Onions', 1],
['Olives', 1], ['Zucchini', 1], ['Pepperoni', 2] ]);
var options = {'title':'Pizza I Ate Last Night',
'width':400, 'height':300};
var chart = new google.visualization.PieChart(document.getElementById
('chart_div'));}
</script>
</head>
<body>
<div id="chart_div"></div></body> </html>
load('visualization','1.0',{'packages':['corechart']});
setOnLoadCallback(drawChart);
函数drawChart(){
var data=new google.visualization.DataTable();
data.addColumn('string','Topping');
data.addColumn('number','Slices');
data.addRows([['蘑菇',3],'洋葱',1],
[橄榄,1],“西葫芦”,1],“意大利香肠”,2]);
var options={'title':'Pizza我昨晚吃的',
“宽度”:400,“高度”:300};
var chart=new google.visualization.PieChart(document.getElementById
('chart_div');}
我的页面将其加载到div的代码:
<html>
<head>
<script type="text/javascript" src="jquery-1.7.1.min.js"></script>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script>
$(document).ready( function () { $('#page').load('in.php');});
</script>
</head>
<body>
<div id=page style="width: 900px; height: 500px;" ></div>
</body>
</html>
$(document.ready(函数(){$('#page').load('in.php');});
单独使用in.php可以很好地工作,但不使用out.php,而loading out.php将在源代码中显示空白页和错误。提前感谢 它会抛出什么错误?它会抛出什么错误?