image doesn't display from database (MySQL/php)

Question

Everything is coming from a PDO and is well displaying with the exception of the image. I wrote the folder's path in the DB as if it was written after a src=

As a beginner in php, i'm wondering if it's the good way to display a picture

EDIT: images are in jpg, console shows a "404" but with the good path

<?php
try {
    $bdd = new PDO(/*the pdo works*/);
}
catch(Exception $e)
{
        die('Erreur : '.$e->getMessage());
}
$reponse = $bdd->query('SELECT * FROM equipepedagogique');//where the picture is
?>
 <div class="enseignants"><!--grid-->
            <!--case-->
            <?php 
            while ($donnees=$reponse->fetch())
            foreach ($donnees as $donnee) { 
                $img=$donnees['photo'];
                $nom=$donnees['name'];
                $prenom=$donnees['surname'];
                $fonction=$donnees['fonction'];?>
            <div class="enseignant">
                <a href="">
                    <img src="<?php echo $img?>" alt="" class="photoEnseignant">
                </a>
                <h1 style="margin-left:auto;margin-right:auto;"><?= $prenom .' '. $nom ?></h1>
                <p><?= $fonction ?></p>
            </div>
            <?php } ?>
        </div>

Is $img a path or a binary value? if it is a binary value print sample value you read from database in the question. — Majid Hajibaba
– Majid Hajibaba, Commented Jun 24, 2021 at 9:28
$img is supposed to display the path of the picture in the folder from the database, stocked with a TEXT type. — Eric Minel
– Eric Minel, Commented Jun 24, 2021 at 9:36

user16304937 · Accepted Answer · 2021-06-24 09:40:17Z

0

<?php
try {
    $bdd = new PDO(/*the pdo works*/);
}
catch(Exception $e)
{
        die('Erreur : '.$e->getMessage());
}
$reponse = $bdd->query('SELECT * FROM equipepedagogique');//where the picture is
?>
 <div class="enseignants"><!--grid-->
            <!--case-->
            <?php 
            ***while ($donnees=$reponse->fetch())*** **//while($donnees=$reponse->mysql_fetch_array())**
            foreach ($donnees as $donnee) { 
                $img=$donnees['photo'];
                $nom=$donnees['name'];
                $prenom=$donnees['surname'];
                $fonction=$donnees['fonction'];?>
            <div class="enseignant">
                <a href="">
                    <img src="<?php echo $img?>" alt="" class="photoEnseignant">
                </a>
                <h1 style="margin-left:auto;margin-right:auto;"><?= $prenom .' '. $nom ?></h1>
                <p><?= $fonction ?></p>
            </div>
            <?php } ?>
        </div>

answered Jun 24, 2021 at 9:40

user16304937

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1 Comment

Eric Minel Over a year ago

thanks , but it doesn't work. but $nom, $prenom and $fonction are displayed

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