Question

m having a problem while submitting the form using jquery & Ajax. my prob is that i want to insert some values into the database before submitting the form. code that i am using is as below for form submission :

$('.buttons').click(function(){

var account_num = $("#accountNum").val();
var ref_num = $("#refNum").val();

var dataString = "account_num="+ escape(account_num) + "&reference_no=" + escape(ref_num);

$("#frmTransaction").hide();
$("#loader_div").show();

$.ajax({
   type : "Post",
   url  : "validate-data.php",
   data : dataString,
   success: function(html) {
      if(html) {
    alert(html);
    $("#frmTransaction").submit();

      } else {
          alert('Unknown Error Please...Retry');    
       return false;
       }
    }, error :function (){
        return false    
    }, 
       complete: function () {  }

});
 return false;

});

html form is as :

<form  method="post" action="https://sample.com/pay" name="frmTransaction" id="frmTransaction">
<input name="accountNum" id="accountNum" type="hidden" size="60" value="<? echo $_POST['account_id'] ?>">
<input name="refNum" id="refNum" type="hidden" size="60" value="<? echo $reference_no;?>" />
<input type="text" class="field" name="name" id="name" />
<input type="text" class="field" name="city" id="city" />
<input type="text" class="field" name="state" id="state" />
<input type="text" class="field" name="postal_code" id="postal_code" />
<input type="submit" class="buttons" name="submitted" value="Proceed" />
</form>

content of validate-data.php is as :

<?PHP
        $account_num = $_REQUEST['account_num'];
    $reference_no = $_REQUEST['reference_no'];

    $countF = mysql_num_rows(mysql_query("SELECT * FROM orders where order_id='".$reference_no."'"));

    if($countF == 0 ) {
        $res = mysql_query("insert into orders(order_id, user_account_num)values('".$reference_no."', '".$account_num."')");
    } 

if($res) 
    echo "$res ";
else 
    echo "";    
?>

Answer 1

only two problem :
One, see your callback success:

success: function(html) {
  if(html) {
  alert(html);
  $("#frmTransaction").submit();

  } else {
      alert('Unknown Error Please...Retry');    
   return false;
   }
}

your if statement will ask to computer, if isset variable html
and two, you print the mysql query on your php code …

if($res) 
   echo "$res ";
else 
   echo "";

and the workaround:
change your jquery code with this one:

$.ajax({
 type : "Post",
 url  : "validate-data.php",
 data : dataString,
 success: function(html) {
 var callback = $.parseJSON(html); // Parse your callback
   if(callback.process == 'success'){
      $("#frmTransaction").submit();
   } else {
      alert('Unknown Error Please...Retry');    
   }
 }, error:function(){
    alert('Unknown Error Please...Retry');    
 }, 
});

and the php code:

<?PHP
 $account_num = mysql_real_escape_string($_POST['account_num']); // make sure `$_POST` and `mysql_real_escape_string` used to prevent sql injection
 $reference_no = mysql_real_escape_string($_POST['reference_no']); // And this line too

 $countF = mysql_num_rows(mysql_query("SELECT * FROM orders where   order_id='".$reference_no."'"));

if($countF == 0 ) {
    $res = mysql_query("insert into orders(order_id,   user_account_num)values('".$reference_no."', '".$account_num."')");
} 

if($res){
  echo json_encode(Array("process" => "success"));
}
else {
  echo json_encode(Array("process" => "fail")); 
}
?>

Good luck, Friend