John Davidson

function checkstock() {



  var stockid = $(".stockid").val();

  var idp = $("#idp").val();

  var transfer = [stockid, idp];

  $.ajax({

    type: "post",

    url: "<?php echo base_url(); ?>index.php/homecontroller/checkstock",

    data: {

      result: JSON.stringify(transfer)

    },

    success: function(response) {

      if (response) {

        $('#msg').html('<span style="color: green;">ONLY ' + stockid + "</span>");



      }

    }

  });

}

<input id="idp" name="prodid" value="<?php echo $product->id;?>" type="hidden">

<input class="stockid" type="radio" onclick="checkstock();" name="size" value="xs">

<input class="stockid" type="radio" onclick="checkstock();" name="size" value="s">

<input class="stockid" type="radio" onclick="checkstock();" name="size" value="m">

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My Answer:
("#idp").val();__ $.ajax({__ url: "",__ type: "POST",__ data: {stockid: stockid, idp: idp},__ success: function(data) {__ $(".stockleft").html(data);__ }__ });__}

In your PHP controller, you can create a function to get the stock value from the database and return it to the AJAX call:

public function get_stock() {__ $stockid = $this->input->post('stockid');__ $idp = $this->input->post('idp');__ $stock = $this->your_model->get_stock($stockid, $idp);__ echo $stock;__}

In your model, you can create a function to get the stock value from the database:

public function get_stock($stockid, $idp) {__ $this->db->select('stock');__ $this->db->where('stockid', $stockid);__ $this->db->where('idp', $idp);__ $query = $this->db->get('products');__ $result = $query->row();__ return $result->stock;__}

Make sure to replace 'your_model' with the actual model name you are using in your CodeIgniter application.