Passing PHP variable through javascript and back again

Question

I have a form that generates lots and lots of rows. Each row has an "Add Notes" button like this

<button onclick=\"myFunction()\">Add Note</button>

and it triggers a popup input through this snippet

<script language="JavaScript" type="text/javascript">
    function myFunction() {
        var x;
        var note = prompt("Customer Note","Write your customer note here...");

        if (note != null) {
            document.getElementById("notes").value = note;
            document.getElementById("notesForm").submit();
        } 
    else{
       return false;
        }
    }
</script>

and submits a form through this section

<form action=\"http://calls.fantomworks.com/functions/notes.php\" 
id='notesForm' name='notesForm' method='post'>
    <input type='hidden' id='ID' name='ID' value='{$row['ID']}' />
    <input type='hidden' id='notes' name='notes' />
    </form>

The problem is that the note is getting passed to the top row instead of the correct {$row['ID']}. How do I pass the {$row['ID']} through this popup and back to the form so that it will be gotten in the notes processor below correctly??

$notesID = $_POST['ID'];
$note = $_POST['notes'];
$note = mysql_real_escape_string($note);
$date= date('Y-m-d');

$result = mysql_query("UPDATE Project_Submissions SET 
                       Notes=CONCAT(Notes,'<br />".$date." ".$note."') 
                       WHERE ID ='".$notesID."'");

I am so lost and could really use some help here. Thank you so much in advance!!

You are vulnerable to sql injection attacks. Escaping $note is NOT enough. you have to escape ALL external data. — Marc B
– Marc B, Commented Jul 31, 2015 at 21:13
As you are new2programming please use the MYSQLI or PDO extensions for you database access. the mysql extension will soon disappear as it is deprecated. Dont waste your time with it. Learn one of the others, please read stackoverflow.com/questions/12859942/… — RiggsFolly
– RiggsFolly, Commented Jul 31, 2015 at 21:56
Dont actually understand what youare trying to say, please look at your question with the tought... they are not clarevoyant and the cannot see over my shoulder — RiggsFolly
– RiggsFolly, Commented Jul 31, 2015 at 21:59

Marc B · Accepted Answer · 2015-07-31 21:59:03Z

1

You need to include WHICH row you're working on in, e.g.

while(...) { 
   <button onclick="myFunction(<?php echo $id ?>);">....</button>
}

and then use that function parameter when you do the ajax call.

answered Jul 31, 2015 at 21:59

Marc B

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1 Comment

new2programming Over a year ago

THAT DID THE TRICK! THANK YOU! <button onclick=\"myFunction('{$row['ID']}')\">Add Note</button> <script language="JavaScript" type="text/javascript"> function myFunction(ID) { var x; var ID = ID; var note = prompt("Customer Note","Write your customer note here..."); if (note != null) { document.getElementById("notes").value = note; document.getElementById("notesID").value = ID; document.getElementById("notesForm").submit(); } else{ return false; } } </script>

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