如何根据点击按钮重新加载我的谷歌图表

这是我用来显示图表和票数的代码。它运行得很好。

在这段代码中，当我点击刷新按钮时，投票会被更新而不刷新页面。我该怎么做？

<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);

}