通过表单传递多个变量

Question

I have the following code, the variable 'name' and 'project' need to go to session variables on submit, in this case the submit happens through an 'onchange'event. The 'name' variable is POSTed but not the 'project' variable. Any ideas where my issue is?

<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<?php
$query = $conn->prepare("SELECT name, project FROM models.models where models.active = 'yes'");
$query->execute();
while($r = $query->fetch(PDO::FETCH_OBJ)){

echo "
<div class='swiper-slide'>
  <input type='image' src='models/thumbs/",$r->name,".jpg' id='name' name='name' value='",$r->name,"' onchange='this.form.submit();'/>
  <input type='hidden' value='",$r->project,"' name='project' />
</div>

";
}

?>
</form>

Answer 1

<?php
$query = $conn->prepare("SELECT name, project FROM models.models where models.active = 'yes'");

$query->execute();
while($r = $query->fetch(PDO::FETCH_OBJ)){

echo "
<div class='swiper-slide'>
  <input type='image' src='models/thumbs/",$r->name,".jpg' id='name' name='name[]' value='",$r->name,"' onchange='this.form.submit();'/>
  <input type='hidden' value='",$r->project,"' name='project' />
</div>

";

}

?>

How to access:

$names = $_POST['name'];

for($i=0;$i<count($names);$i++)
{
   echo $names[$i]."<br>";
}