PHP Checkbox Form Validation with SQL - php

Question

I have mySQL table containing rows of user reviews of films. All columns a are functional except for the 'liked' column. The 'liked' column is a boolean value.
This
The table must be displayed on my website, converting the boolean to a 'yes' or 'no'. Here is the code for the table:
while($review = $reviews->fetch_object("Review")) {
$liked = $review->liked;
$convert = ($liked) ? 'yes' : 'no' ;
echo "<tr> <td>{$review->reviewer} </td> ";
echo "<td>{$review->comment} </td>";
echo "<td> $convert </td></tr>";
As you can see the conversion is functional.
The page also contains a form where users can submit their reviews. Using a checkbox, users 'check' if they like the film, or don't check if they don't.
<div class="form-group">
<label for="liked" class="col-xs-2 c">Did you like this film?:</label>
<div class="col-xs-10">
<input name="liked" type="checkbox" value="1" > Tick yes if you did
</div>
</div>
In a seperate PHP file; 'Process-review', I use an INSERT query to insert new rows into the table.
if(isset($_POST['name']) && isset($_POST['comment']) && isset($_POST['film_id']))
{
$reviewer = $_POST['name'];
$comment = $_POST['comment'];
$film_id = $_POST['film_id'];
if(isset($_POST['liked']))
{
$liked = $_POST['1'];
}
else
{
$liked = $_POST['0'];
}
$db->query("INSERT INTO review (film_id, reviewer, liked, comment) VALUES('$film_id', '$reviewer', '$liked', '$comment')");
header('Location: show-film.php?id='.$film_id);
}
else{
$name = null;
echo "no name supplied";
}
All other fields work except for the 'liked' field. As you can see I have tried using an if statement within the initial if statement to return a '1' if the checkbox is ticked and '0' if the checkbox isn't ticked. Yet when I check the box, the table still returns a 'no'. Even if i change the $convert so both values are 'yes' it still returns no which begs the question where is it getting the 'no' from.
Here is an image of the table to give you a visual representation of what I'm working with here

Answer 1

You're trying to insert values that don't exist.
Replace your if condition with:
if(isset($_POST['liked']) && $_POST['liked'] == 1)
{
$liked = 1;
}
else
{
$liked = 0;
}
Aside from that, you're opening yourself up to an injection attack.
You should use prepared statements.
At the very least run every variable you're passing to the database through mysqli_real_escape_string.