Store values in database table using php explode function - php

Question

I have a form with 4 or more checkboxes.The form is created by a while loop and has data from database table.Each checkbox represents a gift and each gift has a unique id (id1=1, id2=2, id3=3, ...) in database.So user can choose one or more gifts and click the submit button to store his choice.So I need php implode() function to get the string of what choices user made.For example if he chooses the 1st and the 2nd checkbox/gift the value that must be stored should be 1, 2.Instead of this I always have different stored data and don't match with user choices..Any ideas?
Form :
<form method='post' action='insert.php'>
.
.
.
while($row_select4 = $stmt_select4->fetch(PDO::FETCH_ASSOC)){
if($form_points>=$row_select4['points']) {
$points = $row_select4['points'];
echo '
<div>
<div class="feed-activity-list"><div style="border: 0.5px solid green; border-right-style:none;" class="input-group m-b"><span class="input-group-addon">
<input type="checkbox" onclick="enable_form();" id="checkbox" name="opt[]" value="'.$points.'"></span>
<input type="hidden" name="opt2[]" value="'.$row_select4['id'].'">
<div class="feed-element">
<a href="profile.html" class="pull-left">
<img alt="image" class="img-circle" src="'. $row_select4['image_url']. '">
</a>';
?>
<button type="submit" id="submit" name="eksasrgirwsh" class="btn btn-w-m btn-primary">ΕΞΑΡΓΥΡΩΣΗ</button>
</form>
Also (before insertion into table) :
if(isset($_POST['opt2'])){
foreach ($_POST['opt2'] as $value) {
$gift = $_POST['opt2'];
$sliced = array_slice($gift, 0, -1);
$gift_id = implode(", ", $sliced);
}
This is what I use to get the chosen gift_ids and create the string of them
Form updated:
<input type="hidden" disabled="disabled" name="opt2[]" value="'.$row_select4['id'].'">
Code for getting gift_ids:
foreach ($_POST['opt2'] as $value) {
$gift = $_POST['opt2'];
$gift_id = implode(", ", $gift);
}

Answer 1

No need for array_slice
in your php code:
$gift = $_POST['opt2'];
$gift_id = implode(", ", $gift);
By using this, you can get the result you wish.
Edit:
The reason you get all 1,2,3,4 is that you use hidden for gift id.
Try to add disabled="disabled" to the hidden tag which you dont want send to server. After this, you can get correct ids using $_POST['opt2']