我必须显示谷歌图表根据下拉值,其中包含店铺ID 我从MySQL检索数据,没有问题的价值观,我检索数据根据商店ID从ajax,并在输入框中确认它也很好。更新谷歌图表动态下拉使用AJAX和PHP
但我不知道如何用这些值更新该图表,无需重新加载页面。 这里是我的谷歌图表代码与硬编码值。
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>newChart</title>
<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()
{var data = google.visualization.arrayToDataTable([['Company & Model', 'Views'],['Samsung Hero Music E1232B',5],['Samsung Galaxy Y S5360',7],['Samsung Galaxy Ace S5830',7],['Karbonn K 1212',2],]);
var options = {
title: 'Most Popular Item ',
hAxis: {title: 'Brand', titleTextStyle: {color: 'red'}}};
var chart = new google.visualization.ColumnChart(document.getElementById('MPI_chart_div'));
chart.draw(data, options);
}
</script>
</head>
<body>
<h3>COLUMN CHART FOR MOST POPULAR ITEM </h3>
Select Shop <select id="MPI_selected_shop" onchange="MPI_set_shop(this.value);">
<option value="all_Shops">All Shops</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
<input type="text" id="sd" />
<div id="MPI_chart_div" style="width: 800px; height: 400px;"></div>
</body>
</html>
这里是脚本标签内的同一页我的Ajax代码
var xmlHttp
function MPI_set_shop(str)
{
alert(str);
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("Your browser does not support AJAX!");
return;
}
var url="chart.php";
url=url+"?q="+str;
alert(url);
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
function stateChanged()
{
if (xmlHttp.readyState==4)
{
document.getElementById("sd").value=xmlHttp.responseText;
$st=xmlHttp.responseText;
alert($st);
}
}
这里是我在哪里从MySQL使用AJAX
<?php
$testid=0;
$testid=$_REQUEST["q"];
//echo $testid;
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
// Select Database
mysql_select_db("mysql", $con) or die('Could not connect: ' . mysql_error());;
$qMostPopularItem = "SELECT t.pr_id,p.pdt_company_name,p.pdt_model_name,SUM(t.count) AS count FROM tmp AS t INNER JOIN product_mapping AS p ON t.pr_id = p.pr_id AND t.shop_id =$testid GROUP BY pr_id ORDER BY t.count DESC;";
$mpi = mysql_query($qMostPopularItem,$con) or die('Could not fetch MPI: ' . mysql_error());
while($infoMPISW = mysql_fetch_assoc($mpi))
{
echo "['".$infoMPISW['pdt_company_name']." ";
echo $infoMPISW['pdt_model_name'] ."',";
echo $infoMPISW['count'],"],";
}
?>
谢谢你回复卡迈勒。它为我工作。 – Pramod
上述代码中的数据是硬编码的,它工作正常,但我从数据库中获取数据并将其存储在变量中,并且该变量本身的格式(如我所愿)意味着添加方括号和所有这些。但不是“显示变量的值,而是显示变量名称,我应该如何解决它。 – Pramod
首先,我的anwser帮助过你,但到目前为止,你正在传递一个字符串给ajax响应。数组本身,您可以使用json来获取数据并作为数组传递。 – Kamal