error in updating mysql with $.post using jquery - php

Question

i wanna have like for my blog posts, it should work like this: user will click on something and it increase that number by +1 and store it in data base, i have a column named post_like in my db. but after increase 0 to 1 (when i try to increase from 1 to 2 or more) i get error.
jquery:
$("#insert_like").click(function(e){
alert('s')
var like = $("#insert_like").val();
like += 1;
var post_id = $("#post_id").val();
$.post("./inc/like.php", {
like: like,
post_id: post_id
}, function(data, status){
$("#insert_like").text(data);
like = 0;
});
});
php:
<?php
if (isset($_POST['like'])) {
require_once 'db.inc.php';
$like = $_POST['like'];
$post_id = $_POST['post_id'];
$q = "UPDATE posts set post_like = ? WHERE post_id=? LIMIT 1";
$stmt = $conn->prepare($q);
$stmt->bind_param('ii', $like, $post_id);
$stmt->execute();
if ($stmt->affected_rows == 1) {
echo "$like";
} else {
echo "error: $stmt->error";
}
$stmt->close();
$conn->close();
} else {
header('Location: ../home.php');
}
html:
<p>Post like: <span id="insert_like" style="cursor: pointer"><?php echo $post_like ?></span> </p>

Answer 1

You can pass the Post Id from javascript and update the likes in the backend. Consider below example:
$("#insert_like").click(function(e){
$.post("./inc/like.php", {
post_id: $("#post_id").val()
}, function(data, status){
$("#insert_like").text(data);
like = 0;
});
});
and in the backend
<?php
if (isset($_POST['like'])) {
require_once 'db.inc.php';
$post_id = $_POST['post_id'];
$q = "UPDATE posts SET post_like = (post_like + 1) WHERE post_id = ?";
$stmt = $conn->prepare($q);
$stmt->bind_param('i', $post_id);
$stmt->execute();
if ($stmt->affected_rows == 1) {
// get the updated likes and return as response.
} else {
echo "error: $stmt->error";
}
$stmt->close();
$conn->close();
} else {
header('Location: ../home.php');
}
Hope this helps.