I have to make everything happened on the same page. I have used action="<?PHP echo $_SERVER['PHP_SELF']; ?>" here but it is not working. I have insert the PHP query below the form. Basically, my question is how do I make sure the form is posting the values on the same page. If it is updated, a pop up will come up. 
$user_id=$_SESSION['user_id'];
$date = date("l jS \of F Y h:i:s A");
$query1 ="SELECT daily_limit FROM user WHERE user_id='$user_id'"; 
$result1 = mysqli_query($link, $query1) or die(mysqli_error($link));
while ($row1 = mysqli_fetch_array($result1)) {
    $dailylimit=$row1['daily_limit'];
}               
$query2 = "SELECT SUM(debit) AS debited_today FROM transaction WHERE user_id = '$user_id' AND date = CURRENT_DATE" ;
$result2 = mysqli_query($link, $query2) or die (msqli_error($link));
while ($row2 = mysqli_fetch_array($result2)){
    $debited_today = $row2['debited_today'];
}
// form
    <form method="POST" action="<?PHP echo $_SERVER['PHP_SELF']; ?>" > 
        <table id="table">
            <tr>
                <td class="alt">Existing Daily Limit</td> 
                <td>S$ <?php echo $dailylimit; ?> </td>
                <input type="hidden" name="dailylimit" value="<?php echo $dailylimit ?> "/>
            </tr> 
            <tr>
                <td class="alt"><label for="newdailylimit">New Daily Limit</label></td>
                <td>$ <select name="newdailylimit">
                        <option value="100.00">100.00</option>
                        <option value="500.00">500.00</option>
                        <option value="1000.00">1000.00</option>
                        <option value=5000.00">5000.00</option>
                    </select></td>
            </tr>
            <tr>
                <td class="alt">Amount Debited Today</td>
                <td>S$ <?php echo $debited_today; ?></td>
            </tr>
            <tr>
                <td class="alt">Amount Debited Left</td>
                <td>S$ <?php echo ($dailylimit - $debited_today); ?> </td>
            </tr>
        </table>
        <br/>
        <input type="submit" name="submit "value="Submit"></input>
    </form>
// Values I need to POST
$dailylimit = $_POST['dailylimit'];
$newdailylimit = $_POST['newdailylimit'];
if ($dailylimit != $newdailylimit){
    $query = "UPDATE user SET daily_limit='$newdailylimit' WHERE user_id='$user_id'";
    $result = mysqli_query($link, $query) or die(mysqli_error($link));
    echo "<script>alert('You have successfully updated your daily limit');</script>";
}
else if ($dailylimit == $newdailylimit){
    echo "<script>alert('You have selected the same daily limit as your previous one. Please choose a different one. ');</script>";
}
else{
}
 
     
     
    