我正在尝试将mysql查询后的php中的json数据传递给加载google chart api的javascript。查询成功,但数据格式似乎不正确。我在这里做错了什么?我把它保存为.html文件,然后通过浏览器打开它。希望应该这样。
< ?php
$DB_hostname = "localhost";
$DB_Name = "root";
$DB_pass = "root";
$tbl_name="tblname"; // Table name
$con = mysql_connect($DB_Hostname,$DB_Name,$DB_pass) or die(mysql_error());
mysql_select_db("dbname", $con);
$sql="SELECT * FROM $tbl_name WHERE is86 = 1";
$result=mysql_query($sql);
$count1 = mysql_num_rows($result);
$sql="SELECT * FROM $tbl_name WHERE is86 = 0";
$result=mysql_query($sql);
$count2 = mysql_num_rows($result);
$data[0] = array("Changes","count1");
$data[1] = array("Apps with IA Architecture",.$count1.);
$data[2] = array("Apps with no IA Architecture",.$count2.);
$data = json_encode($data);
? >
<html>
<head>
<!--Load the AJAX API-->
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script type="text/javascript">
// Load the Visualization API and the piechart package.
google.load('visualization', '1', {'packages':['corechart']});
// Set a callback to run when the Google Visualization API is loaded.
google.setOnLoadCallback(drawChart);
function drawChart() {
var data = new google.visualization.DataTable(<?=$data?>);
var options = {
title: 'Architecture Changes',
is3D: 'true',
width: 800,
height: 600
};
var chart = new google.visualization.PieChart(document.getElementById('chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>
<!--Div that will hold the pie chart-->
<div id="chart_div"></div>
</body>
</html>
答案 0 :(得分:1)
您使用DataTable构造函数的数据格式不正确,但您应该能够使用arrayToDataTable
方法:
var data = google.visualization.arrayToDataTable(<?=$data?>);
此外,在将.
变量添加到数组时,不应在$count
变量之前和之后包含$data[0] = array("Changes", "count1");
$data[1] = array("Apps with IA Architecture", $count1);
$data[2] = array("Apps with no IA Architecture", $count2);
:
{{1}}
答案 1 :(得分:0)
报价?
var data = new google.visualization.DataTable('<?=$data?>');