Refresh value from MySQL db on php page after clicking a submit button

Question

So all i need to do is refresh a variable displayed on a php page which is stored in a MySQL db. This value is an int which is subtracted by 1 everytime the submit button from a form is clicked. As i've opted to use AJAX to post the form the page isn't being refreshed, therefore the value isn't being updated along with the form submission.

$qry = mysql_query("SELECT codes_remaining FROM users WHERE email= '".$_SESSION['email']."'");
while($row = mysql_fetch_array($qry)) {
    if ($row['codes_remaining'] ==1 ) 
    {
    echo "You have ".$row['codes_remaining'].' code remaining';
    }
    else {
echo "You have ".$row['codes_remaining'].' codes remaining';
}
}

So this code just displays how many "codes" a person has left. I need this value to be refreshed once the submit button has been clicked from the form on the same page.

I'm using the following JavaScript to not refresh the page.

$("#form-submit").click(function(e) { 
        e.preventDefault();
        $.ajax({
        cache: true,
        type: 'POST',
        url: 'process-register.php',
        data: $("#form-register").serialize(),
        success: function(response) {
            $("#output-div").html(response);
        }
        });
    });

Thanks, LS

Your PHP code appears to be vulnerable to SQL injection attacks. — Mark Byers
– Mark Byers, Commented Sep 17, 2011 at 22:24
Well, I need the value to be refreshed once a submit button has been clicked. As I said, I'm using AJAX to process the form so at the moment the page doesn't refresh therefore the value doesn't refresh.. — jcrowson
– jcrowson, Commented Sep 17, 2011 at 22:41

Korvin Szanto · Accepted Answer · 2011-09-17 23:03:30Z

1

If you'd like to update the value, do it like this (jQuery is easiest):

$(".submit").click(function(event) { 
     event.preventDefault();
     $(this).load('file.php',function(val){
        $('#output').text(val);
     });
 });

And in file.php:

<?php
    connect_to_db();
    $returned = get_info_from_db();
    echo $returned;
?>

The jQuery will grab the info on file.php and put it into #output.

answered Sep 17, 2011 at 23:03

Korvin Szanto

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Comments

axel22 · Accepted Answer · 2011-09-18 21:55:48Z

1

Maybe It's just me, but why not use jQuery .load function?

    $("#form-submit").click(function(e) { 
        e.preventDefault();
        $(this).load('process-register.php');
    });

Maybe not ethical nor the correct way of doing this but everytime you click on #form-submit, it loads that file and therefore processes it everytime. Also note that if you load a file that uses MySQL connection yes has no mysql_connect or mysql_select_db configured, it obviously won't work. I've had that for quite some times.

edited Sep 18, 2011 at 21:55

axel22

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answered Sep 17, 2011 at 22:41

That Other Guy

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Comments

Fata1Err0r · Accepted Answer · 2015-02-11 17:06:52Z

0

In your 'success', you could possibly just throw in

$("#WhereYouWantTheOutput").load("process-register.php");

That way whenever your submit succeeds, it'll also load the output for you. Just replace #WhereYouWantTheOutput with the name of where you want the output placed.

answered Feb 11, 2015 at 17:06

Fata1Err0r

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