0

I want to draw a pie chart where value will be taken from MySQL database. But it is not working. But if I give the manual value then the pie chart is showing. Below is my code:

<?php
include "libchart/classes/libchart.php";

header("Content-type: image/png");

$chart = new PieChart(500, 260);
$con=mysqli_connect("localhost","root","","bkash");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT count(*) FROM dialer_rate where mno='tnr' and success=1 ");
$result1 = mysqli_query($con,"SELECT count(*) FROM dialer_rate where mno='tnr' and failed=1");
$dataSet = new XYDataSet();
$dataSet->addPoint(new Point("Success", $result));
$dataSet->addPoint(new Point("Failed", $result1));
#$dataSet->addPoint(new Point(" (50)", 50));
$chart->setDataSet($dataSet);

$chart->setTitle("bKash USSD Dialer Success/Fail rate");
$chart->render();
?>

But if I give the manual value in below two fields then it is working. Anyone please help.

$dataSet->addPoint(new Point("Success", 20));
$dataSet->addPoint(new Point("Failed", 80));
Rodrigo Taboada
  • 2,727
  • 4
  • 24
  • 27
Basudev
  • 155
  • 5
  • 13

1 Answers1

0

try this 

...
$result = mysqli_query($con,"SELECT count(*) as count FROM dialer_rate where mno='tnr' and success=1 ");
$result1 = mysqli_query($con,"SELECT count(*) as count1 FROM dialer_rate where mno='tnr' and failed=1");
$result = $result->fetch_object();
$result1 = $result1->fetch_object();
$dataSet = new XYDataSet();
$dataSet->addPoint(new Point("Success", $result->count));
$dataSet->addPoint(new Point("Failed", $result1->count1));
...
Balaji
  • 850
  • 7
  • 15