Validate input value against calculated value

Question

Hi Chaps,
Need a bit of guidence with some PHP code.
I have a Query that estimates a quote ($price_total) for a job.
The estimate ($price_total) is the value of an input (jobquote), and the database is updated once the form is submitted (using a seperate script.php page).
What I need, is to validate the entered value of 'jobquote' against the estimated value of $price_total, just incase a 'custom' price has been agreed with a customer.
If the values are different, then I need an 'admin override' radio button (admin_quote enum('y','n')) to appear.
If someone can help or point me in the right direction, I'd be most grateful.
Cheers

Answer 1

Got a bit further:
PHP Code:
<input type='text' name='jobquote' value="<?php echo $price_total; ?>"/>
        <input type='hidden' name='original_jobquote' value="<?php echo $price_total; ?>"/>
        <?php
if ($_POST['original_jobquote'] != $_POST['jobquote'])
{ ?>
<span id="spryradio1">
<input type="radio" name="jobquoteadmin" value="y" id="radio" />Confirm<br />
<span class="radioRequiredMsg">Please confirm Admin Override</span></span>
<?php };
?>
Problem 1. The information is '$_POST'ed to a script file, not to the page itself and at the moment, the $_POST takes place before the PHP validation takes place. If I remove the link to the script page, the validation works.
Problem 2. (link to script removed for testing) If I change the value (from the default 'original_quote'), then submit, the page reloads, the 'Confirm' radio button appears, but the value of 'job_quote' has reverted back thte default:
1. Start - job_quote = £350
2. Change - job_quote = £100
3. Submit
4. Page reloads - job_quote = £350, confirm appears
Is there a way around this?