如何通过Ajax发送$ _GET

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";

?>

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.

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

Best regards MrKampf

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'];
}