将json数据从PHP传递到javascript - Google Pie Chart

时间:2013-10-24 18:38:08

标签: javascript php mysql google-visualization

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>

2 个答案:

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?>');