enna

Members
  • Content count

    2
  • Joined

  • Last visited

Everything posted by enna

  1. hi,i am trying to get the data from my database and i successfully got what i wanted.. but when i added a link that when clicked should alter the content of the div (where the chart is placed) with a new chart with different data from the database, i only get the word "Chart" inside the div.. can someone enlighten me? thank you! heres a snippet of the codes: <script> function openProvList(code){ var welcome= document.getElementById("graphs"); $.ajax({ type: "POST", url: "obpims.php", data: "code="+ code, dataType:"html", success: function (html) { welcome.innerHTML = html; } }); } </script> <?php include('../dbconnect.php'); include("FusionCharts.php"); ?> <div id="graphs"> <a href="#" onClick="openProvList(<?php echo $rowProv['province_code']; ?>)">CLICK ME</a> <div class="gen-chart-render"> <?php $strXML1 = "<chart caption='ToTal Barangay Projects' pieSliceDepth='30' showBorder='1' formatNumberScale='0' xAxisName='City/Municipality' yAxisName='Projects' numberSuffix=' projects'>"; $strQuery1 = "select * from province"; $result1 = mysql_query($strQuery1) or die(mysql_error()); if ($result1) { while($ors1 = mysql_fetch_array($result1)) { $str1 = "select COUNT(*) from project a, brgy b, city c, province d where a.brgy_code=b.brgy_code and b.city_code=c.city_code and c.province_code=". $ors1['province_code']; $result12 = mysql_query($str1) or die(mysql_error()); if($result12){ $cnt1 = mysql_fetch_row($result12); } else{ echo "Error in query: ".mysql_error();} $strXML1 .= "<set label='" . $ors1['province_name'] . "' value='" . $cnt1[0] . "' />"; mysql_free_result($result12); } } $strXML1 .= "</chart>"; echo renderChart("../tmp/FusionCharts/Column3D.swf", "", $strXML1, "myChartId2", 600, 260, false, false); ?> </div> </div> inside obpims.php that should replace the content of div (id='graphs') <?php include('../dbconnect.php'); include("FusionCharts.php"); $code = htmlspecialchars(trim($_POST['code'])); ?> <div class="gen-chart-render"> <?php $strXML = "<chart caption='ToTal Allocated Budget' pieSliceDepth='30' showBorder='1' formatNumberScale='0' numberPrefix='Php '>"; $strQuery = "select * from city"; $result = mysql_query($strQuery) or die(mysql_error()); if ($result) { while($ors = mysql_fetch_array($result)) { $str = "select sum(cost) as totalCost from project a, brgy b, city c where a.brgy_code=b.brgy_code and b.city_code=". $ors['city_code']; $result2 = mysql_query($str) or die(mysql_error()); $cnt = mysql_fetch_array($result2); $strXML .= "<set label='" . $ors['city_name'] . "' value='" . $cnt[totalCost] . "' />"; mysql_free_result($result2); } } $strXML .= "</chart>"; echo renderChart("Pie3D.swf", "", $strXML, "myChartId4", 600, 260, false, false); ?> </div>
  2. Problem Generating Chart

    i have found the solution.. i just need to change renderChart("Pie3D.swf", "", $strXML, "myChartId4", 600, 260, false, false); to renderChartHTML("Pie3D.swf", "", $strXML, "myChartId4", 600, 260, false, false); *cheers!*