我希望有人可以帮助我在这里,我一直在寻找highcharts作为从我的sql数据库中绘制我的动态数据的方式....我来了到目前为止是在下面,这很奇怪,看起来是正确的,数据是正确的,但是,我无法让我的图表呈现,现在我已经尝试通过Chrome浏览器和IE查看图表,但没有结果,可以任何人都可以看到我可能出错的地方,或者出现错误的地方......我的数据传入JS数组,所以渲染结束时出现问题......任何帮助都将非常感谢。我关闭了我的html标签,但由于某种原因,它不显示在这篇文章中...如果有人有任何建议,我会很高兴听到他们!问题渲染Highcharts - 通过PHP/MySQL填充
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src='highcharts.js' type='text/javascript'> </script>
<script src='exporting.js' type='text/javascript'> </script>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"type="text/javascript"></script>
</head>
<body>
<?php
include "connect_db.php";
$SQL1 = "SELECT review.reviewForum AS reviewForum,
COUNT(CASE WHEN mom.result = 'Approved' THEN 'yes' ELSE NULL END) AS Approved,
COUNT(CASE WHEN mom.result = 'NOT Approved' THEN 'yes' ELSE NULL END) AS Not_Approved,
COUNT(CASE WHEN mom.result = 'Cancelled' THEN 'yes' ELSE NULL END) AS Cancelled
FROM review INNER JOIN mom ON mom.reviewId = review.reviewId GROUP BY review.reviewForum";
$result1 = mysql_query($SQL1);
$data1 = array();
while ($row = mysql_fetch_array($result1)) {
$data1[] = $row['reviewForum'];
}
$result2 = mysql_query($SQL1);
$data2 = array();
while ($row = mysql_fetch_array($result2)) {
$data2[] = $row['Approved'];
}
$result3 = mysql_query($SQL1);
$data3 = array();
while ($row = mysql_fetch_array($result3)) {
$data3[] = $row['Not_Approved'];
}
$result4= mysql_query($SQL1);
$data4 = array();
while ($row = mysql_fetch_array($result4)) {
$data4[] = $row['Cancelled'];
}
?>
<script type="text/javascript">
$(document).ready(function() {
var chart = new Highcharts.Chart({
chart: {
renderTo: 'container',
type: 'column'
},
title: {
text: 'TC REVIEW RESULT STATS'
},
xAxis: {
categories: [<?php echo join($data1, ',') ?>],
},
yAxis: {
min:0
},
legend: {
layout: 'vertical',
backgroundColor: '#FFFFFF',
align: 'left',
verticalAlign: 'top',
x: 50,
y: 35,
floating: true,
shadow: true
},
plotOptions: {
column: {
pointPadding: 0.2,
borderWidth: 0
}
},
series: [ {
name: 'Approved',
data: [<?php echo join($data2, ',') ?>],
pointStart: 0
//pointInterval
},
{
name: 'Unapproved',
data: [<?php echo join($data3, ',') ?>],
pointStart: 0
//pointInterval
},
{
name: 'Cancelled',
data: [<?php echo join($data4, ',') ?>],
pointStart: 0
//pointInterval
},
]
});
});
</script>
<div id="container" style="min-width: 400px; height: 400px; margin: 0 auto"></div>
</body>
您的html标记可能会丢失,因为您可能需要在该行之后添加一个额外的输入(以使其成为代码的一部分)。 – EricG 2012-07-13 12:24:50