Problem with query to update foreign key column - php

Question

I am working on a cms for properties/ads in oop php for learning purposes. I have three tables that are connected with pivot table.
photos (id, name, extension),
property_photo (id, property_id, photo_id),
properties (id, title, description, main_photo_id)
I have a gallery of photos for every property and I am trying to be able to insert main photo (one of existing photos in gallery) for each property through foreign key (main_photo_id) and display that photo on a different page. I am having trouble writing function (query) in model. Any help is much appreciated. Here is some of my code:
AdModel:
public function MainPhotoInsert($id)
{
$this->db->query('INSERT INTO properties (main_photo_id) VALUES (:main_photo_id) SELECT id FROM PHOTOS WHERE id = :id LIMIT 1');
$this->db->bind(':id', $id);
$row = $this->db->single();
return $row;
}
AdsController:
public function galleryAction()
{
if (!isset($_GET['id'])) {
$photo_id = $_SESSION['photo_id'];
} else {
$photo_id = $_GET['id'];
}
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$_POST = filter_input_array(INPUT_POST, FILTER_SANITIZE_STRING);
if(isset($_POST['radio']))
{
$this->AdModel->MainPhotoInsert($photo_id);
redirect('ads/index');
}
}
$data = $this->AdModel->getPhotosForProperty($photo_id);
$data1 = $this->AdModel->MainPhotoData($photo_id);
$this->view->render('ads/gallery', $data, $data1);
}
gallery.php:
<form action="/ads/gallery?id=<?php echo $_GET['id']; ?>" method="POST">
<?php foreach ($data as $key => $value) : ?>
<img src="<?php echo '/public/photos/'.$value->name.'.'.$value->extension ?>" class="img-fluid img-thumbnail" width="250" height="250">
<input type="radio" name="radio" value="<?php echo $value->photo_id; ?>" >Make main
<br>
<?php endforeach; ?>
<br>
<br>
<button type="submit" name="submit" value="submit" class="btn btn-success form-control">Submit</button>
</form>

Answer 1

You shouldn't have a select clause in your insert statement (at least not for what you are trying to do). IF you only ever need to set it once then you need to tweak the query to include the other two values (title and description) or they will always be blank. It should end up looking something like this:
INSERT INTO properties (main_photo_id, title, description) VALUES (:main_photo_id, : title, : description)
More likely, you want an upsert (update if a relevant row already exists, insert if one doesn't). In MySQL the syntax is insert ... on duplicate key update. This means you are going to need a primary key on the properties table (it's unclear if you already have one). The syntax is pretty similar to the insert above but without knowing the exact structure of the table I can't give you the exact query.
Update:
The on duplicate key syntax would look something like this (it depends on how you have your primary key set up on the table, e.g. if main_photo_id is the primary key then this likely won't work):
INSERT INTO properties
(id, main_photo_id, title, description)
VALUES
(:id, :main_photo_id, : title, : description)
ON DUPLICATE KEY UPDATE
main_photo_id = :main_photo_id,
title = :title,
description = :description
Side note:
gallery.php is also going to produce a separate HTML form for each image with one radio button on the form which isn't really what you want. You should move the form to wrap around the entire foreach loop so that you have one form. Then you should put the image ID as the value for the radio button.