How to send $_GET by Ajax - php

Question

I am trying to send an $_GET['CategoryID'] trought ajax to call in the destination file getdata.php and I can't make it work, I don't find the perfect info here. I know that I am really noob, but I am trying really hard to learn.
I been trying a lot of different code and it still not working.
<button type="button" name="btn_more" data-vid="<?php echo $stockID; ?>" id="btn_more" class="btn btn-success form-control">Ver Mais</button>
<input class="form-control" id="PresentCategoryID" name="PresentCategoryID" data-cat="<?php echo $_GET['categoryID']; ?>" value="<?php echo $_GET['categoryID'];
<script>
$(document).ready(function(){
$(document).on('click', '#btn_more', function(){
var last_video_id = $(this).data("vid");
var PresentCategoryID= ('PresentCategoryID');
$('#btn_more').html("<div class=loader></div>");
$.ajax({
url:"getdata.php",
method:"POST",
data:{
last_video_id:last_video_id,
PresentCategoryID:PresentCategoryID},
dataType:"text",
success:function(data)
{
if(data != '')
{
$('#remove_row').remove();
$('#result').append(data);
}
else
{
$('#btn_more').html("No Data");
}
}
});
});
});
</script>
My objective it's to call the categoryID in the getdata.php, like this,
<?php
$output = '';
$stockID = '';
$PresentCategoryID = '';
sleep(1);
include 'includes/dbh.inc.php';
include 'includes/rating.inc.php';
$sql = "SELECT stock.stockID, stock.name, stock.marca, stock.origem, stock.categoryID, stock.thumbnail, category.name AS catname FROM stock JOIN category ON stock.categoryID=category.categoryID WHERE stock.categoryID='$PresentCategoryID' AND stockID > ".$_POST['last_video_id']." LIMIT 4";
?>

Answer 1

var PresentCategoryID= ('PresentCategoryID')
should be
var PresentCategoryID= $('#PresentCategoryID').val();
You need to use $ to select the element, add the # prefix to use it as an ID, and .val() to get the value of the input.

Answer 2

hey check this answer which will help your problem
$(document).ready(function() {
$(document).on('click', '#btn_more', function() {
var last_video_id = $(this).data("vid");
var PresentCategoryID = ('#PresentCategoryID').val();
$('#btn_more').html("<div class=loader></div>");
var data = {
last_video_id,
PresentCategoryID
};
$.get('getdata.php', JSON.stringify(data)).done(response => {
console.log(response);
if (response != '') {
$('#remove_row').remove();
$('#result').append(response);
} else {
$('#btn_more').html("No Data");
}
}).fail(() => {
console.log("Something went wrong");
});
});
});
PHP SCRIPT
<? php
include 'includes/dbh.inc.php';
include 'includes/rating.inc.php';
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
//if you using get request
//recommended way to get the data use the mysqlconn->real_escape_string($_GET['last_video_id']);
$last_video_id = $_GET['last_video_id'];
$PresentCategoryID = $_GET['PresentCategoryID'];
sleep(1);
$sql = "SELECT stock.stockID, stock.name, stock.marca, stock.origem, stock.categoryID, stock.thumbnail, category.name AS catname FROM stock JOIN category ON stock.categoryID=category.categoryID WHERE stock.categoryID='$PresentCategoryID' AND stockID='$$last_video_id'";
" LIMIT 4";
} else if ($_SERVER['REQUEST_METHOD'] == 'POST') {
//if you using post request then in jquery remove $.get to just $.post
$data = json_decode(file_get_contents('php://input'),true);
$last_video_id = $data['last_video_id'];
$PresentCategoryID = $data['PresentCategoryID'];
}

Answer 3

you want to send via 'GET' but you use the method 'POST'.
Best regards
MrKampf