Dans ce code ci-dessous je veux générer camembert.et j'ai cette erreur Undefined variable: mysqli, Appel à une fonction de membre query() sur un non-objet à la ligne 29. S'il vous plaît quelqu'un m'aider rectifier le problème pour obtenir une solution.Erreur sur la génération graphique
<!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>
<title>Pie Chart Demo (LibChart)- http://codeofaninja.com/</title>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-15" />
</head>
<body>
<?php
//include the library
include "libchart/libchart/classes/libchart.php";
//new pie chart instance
$chart = new PieChart(500, 300);
//data set instance
$dataSet = new XYDataSet();
//actual data
//get data from the database
//include database connection
include 'db_connect.php';
//query all records from the database
$query = "select * from programming_languages";
//execute the query
$result = $mysqli->query($query);
//get number of rows returned
$num_results = $result->num_rows;
if($num_results > 0){
while($row = $result->fetch_assoc()){
extract($row);
$dataSet->addPoint(new Point("{$name} {$ratings})", $ratings));
}
//finalize dataset
$chart->setDataSet($dataSet);
//set chart title
$chart->setTitle("Tiobe Top Programming Languages for June 2012");
//render as an image and store under "generated" folder
$chart->render("generated/1.png");
//pull the generated chart where it was stored
echo "<img alt='Pie chart' src='generated/1.png' style='border: 1px solid gray;'/>";
}else{
echo "No programming languages found in the database.";
}
?>
</body>
</html>