draw a pie chart using php and mysql

Question

I'm trying to draw a pie chart using PHP and MySQL, but I don't see anything, any error just a blank page, I don't know where is the problem exactly, my query is correct. Any help, any suggestion! Thank you. This is my code that I'm using:

<?php
    $dsn='mysql:host=localhost;dbname=tp3_php';
    $user='root';
    $pass='';
    try {
      $bdd = new PDO($dsn,$user,$pass);
    } catch (Exception $e) {
      die('Erreur : ' . $e->getMessage());
    }    
    $sql="
    SELECT Nom_matiere
         , COUNT(ID_etudiant)  
      FROM note
         , matiere 
     WHERE note>=12 
       and matiere.Num_matiere = note.Num_matiere 
     GROUP 
        BY note.Num_matiere
    ";
            $sth = $bdd->query($sql);
?>
<html>
  <head>
    <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script>

    <script type="text/javascript">
      google.charts.load('current', {'packages':['corechart']});
      google.charts.setOnLoadCallback(drawChart);

      function drawChart() {

        var data = google.visualization.arrayToDataTable([
          ['Nom_matiere', 'taux de reussis'],
          <?php

              while ($result=$sth->fetchAll()) 
              {

      echo "['".$results['Nom_matiere']."',".$results['COUNT(ID_etudiant)']."],";
              }

          ?>

        ]);

        var options = {
          title: 'Taux de réussite des étudiants par module'
        };

        var chart = new google.visualization.PieChart(document.getElementById('piechart'));

        chart.draw(data, options);
      }
    </script>
  </head>
  <body>
    <div id="piechart" style="width: 900px; height: 500px;"></div>
  </body>
</html>

@strawberry i did $sql="SELECT Nom_matiere, COUNT(ID_etudiant) as ID FROM note,matiere WHERE note>=12 and matiere.Num_matiere=note.Num_matiere GROUP BY note.Num_matiere"; then i changed echo "['".$results['Nom_matiere']."',".$results['COUNT(ID_etudiant)']."],"; with echo "['".$results['Nom_matiere']."',".$results['ID']."],"; but nothing changed !! — Fatima Assendal
– Fatima Assendal, Commented May 9, 2020 at 21:16
i told you I rename count(ID_etudiant) , count(ID_etudiant) as ID as you told me but i get the same error @strawberry — Fatima Assendal
– Fatima Assendal, Commented May 9, 2020 at 22:09

kmoser · Accepted Answer · 2020-05-10 02:36:03Z

1

It worked for me: I added this line $result=$sth->fetchAll(); and I added a loop foreach, the code will be like this:

['Nom_matiere', 'taux de reussis'],
          <?php
              $result=$sth->fetchAll();
            foreach ($result as $row) { 
              echo "['".$row['Nom_matiere']."',".$row['COUNT(ID_etudiant)']."],";
            }

          ?>

edited May 10, 2020 at 2:36

kmoser

9,3693 gold badges28 silver badges45 bronze badges

answered May 9, 2020 at 22:19

Fatima Assendal

111 bronze badge

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3 Comments

Strawberry Over a year ago

Cool. FWIW, I find this easier to read: echo "['{$row['Nom_matiere']}',{$row['id']}],";

Strawberry Over a year ago

Also, it doesn't make sense to GROUP BY a different column from the one that is selected - Nom_matiere <> note.num_matiere

Strawberry Over a year ago

And always write your JOINs using explicit JOIN syntax. We stopped using comma joins in 1992. And qualify ALL column names !!

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