这是我用来显示图表和显示投票数的代码。它运作得很好。
在此代码中,当我单击刷新按钮时,将更新投票而不刷新页面。我怎么能这样做?
<div id="barchart" style="width: 100%; height: 200px;margin-top:0px;"></div>
<button>Refresh</button>
<?php
$q_id = $row['q_id'];
include "connection.php";
$sql1 = mysql_query("SELECT * FROM votes where q_id='$q_id'");
$count1 = mysql_num_rows($sql1);
$sql2 = mysql_query("SELECT * FROM votes where q_id='$q_id' and answer='Yes'");
$count2 = mysql_num_rows($sql2);
$sql3 = mysql_query("SELECT * FROM votes where q_id='$q_id' and answer='No'");
$count3 = mysql_num_rows($sql3);
?>
<script type="text/javascript">
google.load("visualization", "1", {packages:["corechart"]});google.setOnLoadCallback(drawChart);
google.setOnLoadCallback(render_applicant_sources);
function drawChart() {
var dataTable = new google.visualization.DataTable();
dataTable.addColumn('string', 'Task');
dataTable.addColumn('number', 'Hours per Day');
// A column for custom tooltip content
dataTable.addColumn({type: 'string', role: 'tooltip'});
dataTable.addRows([
['<?php echo "Yes"; echo " "; echo ' ('.$count2.' '.$vname1.')'; ?>', <?php echo $count2; ?> ,'<?php echo "Yes"; ?>'],
['<?php echo "No"; echo " "; echo ' ('.$count3.' '.$vname2.')'; ?>', <?php echo $count3; ?>,'<?php echo "No"; ?>']
]);
var chart = new google.visualization.PieChart(document.getElementById('barchart'));
var options = {
backgroundColor: 'transparent',
pieSliceText:'none',
chartArea: {top: 10},
titleTextStyle: {bold: false},
enableInteractivity: false,
legend: {position: 'labeled',textStyle: {color: '#000',fontSize:13}},
height:250,
fontName:"'Source Sans Pro', sans-serif",
tooltip: {
text: 'value'
}
};
chart.draw(dataTable, options);
}
答案 0 :(得分:0)
实现这一目标的最常见方式:
如果您以这种方式获取数据,则刷新按钮可以再次执行以下步骤:从服务URL获取新数据并根据该数据重绘图表。这种设计有时被称为面向服务的体系结构(SOA)。