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我正在尝试使用 Google 图表并集成到我的 android 应用程序中。在此之前，我尝试在 Google chrome 中检查我的 html 代码。在该代码中，我尝试集成以从 MYSQL 获取数据。我的代码如下

<?php

$server = "localhost";
    $user="root";
    $password="";  
    $database = "mobiledb";

    $connection = mysql_connect($server,$user,$password)
    or
trigger_error(mysql_error(),E_USER_ERROR);
    $db = mysql_select_db($database,$connection);



$sth = mysql_query("SELECT * FROM `table` WHERE `x-axis` AND `y-axis` LIMIT 0, 30 ");

while($r = mysql_fetch_assoc($sth)) {
$arr2=array_keys($x-axis);
$arr1=array_values($y-axis);
}
for($i=0;$i<count($arr1);$i++)
{
    $chart_array[$i]=array((string)$arr2[$i],intval($arr1[$i]));
}
echo "<pre>";
$data=json_encode($chart_array);
?>

<html>
  <head>


    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript" src="https://www.google.com/jsapi"></script>
    <script type="text/javascript">
      google.load("visualization", "1", {packages:["corechart"]});
      google.setOnLoadCallback(drawChart);
      function drawChart() {

      $(function () {
   var data = new google.visualization.DataTable();
        data.addColumn("string", "x-axis");
        data.addColumn("number", "y-axis");
         data.addRows(<?php $data ?>);
    });         
         var options = {
           title: 'Details',
          is3D: 'true',
          width: 800,
          height: 600
        };

        var chart = new google.visualization.ColumnChart(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>

但我看不到图表。我收到以下错误 "; $data=json_encode($chart_array); ?>当我检查页面时，我发现了以下错误 * Uncaught SyntaxError: Unexpected token < * 请帮我解决这个问题。

4

2 回答 2

0

您可能需要更正这条线

data.addRows(<?php $data ?>); //incorrect 
data.addRows(<?php echo($data); ?>); //you forgot to echo it

PHP你的代码有很多问题

  //what is this?
  $sth = mysql_query("SELECT * FROM `table` WHERE `x-axis` AND `y-axis` LIMIT 0, 30 ");

  //probably you want this?
  $sth = mysql_query("SELECT `x-axis`, `y-axis` FROM `table` LIMIT 0, 30 ");

  //this is wrong 
  while($r = mysql_fetch_assoc($sth)) {
    $arr2=array_keys($x-axis);
    $arr1=array_values($y-axis);
  }

  //try this 
  while($r = mysql_fetch_assoc($sth)) {
    $arr2=array_keys($r['x-axis']);
    $arr1=array_values($r['y-axis']);
  }

 //where  is $chart_array defined? It should be like this 
 $chart_array = array();
 for($i=0;$i<count($arr1);$i++)
 {
     $chart_array[$i]=array((string)$arr2[$i],intval($arr1[$i])); 
 }
于 2013-06-16T13:17:40.427 回答
0

检查您的查询

 $sth = mysql_query("SELECT `x-axis`,`y-axis`  FROM `table`  LIMIT 0, 30 ");

如果你想要列 x 轴和 y 轴然后使用这个

然后纠正这个

  data.addRows(<?php echo $data ?>);
于 2013-06-16T13:17:51.963 回答