How to save to database by using ajax - html

Question

I have a code which is works fine, but the data cannot save to the database. I want to insert cost, currency_rate, profit_rate and pprice to database through Ajax. Here are the code of javascript and update.php, I have tried to modify the code to save in my Mysql server, but it didn't success. Can someone help with this?
javascript
$(".profitRate").change(function() {
var myArray = [];
//find closest table->next table
var elem = $(this).closest('table').next('table');
var action = elem.find('tr').data('action');
console.log(action)
var profitRate = Number($("#profitRate").val());
//looping
elem.find('tr').each(function() {
//get cost
var cost = $(this).find('input[name=cost]').val();
//get curency rate
var currecy_rate = $(this).find('select[name=currency_rate]').val();
//calculate profit
var profit_total = Math.round(cost * profitRate * currecy_rate)
$(this).find('input[name=pprice]').val(profit_total)
//add to json object
auto_array = {};
auto_array["cost"] = cost;
auto_array["currecy_rate"] = currecy_rate;
auto_array["pprice"] = profit_total;
myArray.push(auto_array) //push to array
});
console.log(myArray)
form_data = elem.find('tr').data('action');
$.ajax({
data: {
action: action,
form_data: form_data,
},
url: 'update.php',
type: 'post',
beforeSend: function() {
},
success: function(data) {
if(data == 1){
}
}
});
})
update.php
<?php
if ($_POST['action'] == 'update_price') {
parse_str($_POST['form_data'], $my_form_data);
$id = $my_form_data['id'];
$cost = $my_form_data['cost'];
$profit_rate = $my_form_data['profit_rate'];
$currency_rate = $my_form_data['currency_rate'];
$pprice = $my_form_data['pprice'];
$sql = $query = $finalquery = $sqlresult = '';
if ($cost){
$sql.="cost='$cost',";
}
if ($profit_rate){
$sql.="profit_rate='$profit_rate',";
}
if ($currency_rate){
$sql.="currency_rate='$currency_rate',";
}
if ($pprice){
$sql.="pprice='$pprice',";
$finalquery = rtrim($sql,',');
$query="UPDATE `gp_info` SET $finalquery where id=$id";
$sqlresult=mysql_query($query);
if($sqlresult){
$reback=1;
}else{
$reback=0;
}
echo $reback;
}
}