Last ID is added to DB [duplicate] - php

Question

This question already has an answer here:
Multiple hidden input with same name, always retrieve the last input [duplicate]
(1 answer)
Closed 4 years ago.
index.php
<form method="POST" action="add_to_db.php">
<?php
foreach ($response->getGraphEdge() as $graphNode) :
echo
"<div class='form-check mb-3'>" .
"<input type='radio' name='fb_name' class='form-check-input mt-3' value='".$graphNode['name']."'>" .
"<img class='mx-2' src='" . $graphNode['picture']['url'] . "'>" .
"<label class='form-check-label' for='fb_name'>" . $graphNode['name'] . '</label>' .
"<input type='hidden' name='fb_id' value='" . $graphNode['id'] . "'>" .
"<input type='hidden' name='fb_access_token' value='" . $graphNode['access_token'] . "'>" .
"</div>";
endforeach; ?>
</form>
add_to_db.php
if(isset($_POST['submit'])) {
$query = new db;
$b = $query->Query('SELECT * FROM user WHERE user_ID = 1');
$fb_id = $_POST['fb_id'];
$fb_name = $_POST['fb_name'];
$fb_access_token = $_POST['fb_access_token'];
$update = $query->Query("UPDATE user
SET user_fb_page_id = '$fb_id',
user_fb_page_name = '$fb_name',
user_fb_page_access_token = '$fb_access_token'
WHERE user_ID = 1 ");
if(!$update) {
echo '<div class="alert alert-success" role="alert">Updated Successful!</div>';
} else {
echo '<div class="alert alert-danger" role="alert">Oh no! Something went wrong ☹️</div>';
}
}
Problem:
When a user clicks on the Facebook Page that they want, the value to their name & access token is successfully updated to the database however it'll update the last fb_id to the db which is clearly not what I want! I am not sure at all why the ID does this yet none of the other attributes do it

Answer 1

The form input fields in one iteration of your foreach loop have the same names as the input fields in the other iterations.
This produces a form where all of the input fields that hold an id have the exact same name. (The same goes for the input fields that hold an access token and the radio buttons.) Because the fields have the same name, only the last value of each set is submitted to the server, with exception of the radio button which just submits the selected value.
To fix this, you need to give all of the fields in your form an unique name. I've done so here by adding a $key (alternatively you can use the fb_id, if that's unique).
When the form is submitted, only one value should exist in the POST array for $_POST['key'], this is the key of the selected radio button. You can then use this key to find the id, name and access token that go with it:
<form method="POST" action="add_to_db.php">
<?php
foreach ($response->getGraphEdge() as $key => $graphNode) :
echo
"<div class='form-check mb-3'>" .
"<input type='radio' name='key' class='form-check-input mt-3' value='".$key."'>" .
"<img class='mx-2' src='" . $graphNode['picture']['url'] . "'>" .
"<label class='form-check-label'>" . $graphNode['name'] . '</label>' .
"<input type='hidden' name='fb_name[".$key."]' value='" . $graphNode['name'] . "'>" .
"<input type='hidden' name='fb_id[".$key."]' value='" . $graphNode['id'] . "'>" .
"<input type='hidden' name='fb_access_token[".$key."]' value='" . $graphNode['access_token'] . "'>" .
"</div>";
endforeach; ?>
</form>
and
if(isset($_POST['submit'])) {
$query = new db;
$b = $query->Query('SELECT * FROM user WHERE user_ID = 1');
$key = $_POST['key'];
$fb_id = $_POST['fb_id'][$key];
$fb_name = $_POST['fb_name'][$key];
$fb_access_token = $_POST['fb_access_token'][$key];
$update = $query->Query("UPDATE user
SET user_fb_page_id = '$fb_id',
user_fb_page_name = '$fb_name',
user_fb_page_access_token = '$fb_access_token'
WHERE user_ID = 1 ");
if(!$update) {
echo '<div class="alert alert-success" role="alert">Updated Successful!</div>';
} else {
echo '<div class="alert alert-danger" role="alert">Oh no! Something went wrong ☹️</div>';
}
}
(I've removed the "label for" because it wasn't working. To get it to work you can add id='$key' to the radio input, and label for='$key' to the label.)