kool_samule Posted January 20, 2010 Share Posted January 20, 2010 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 Link to comment https://forums.phpfreaks.com/topic/189150-validate-input-value-against-calculated-value/ Share on other sites More sharing options...
Buddski Posted January 20, 2010 Share Posted January 20, 2010 Im a little confused.. You have said that estimate IS the value of the jobquote input.. Which means they should be the same and then you want to compare them.. Have I missed something? Link to comment https://forums.phpfreaks.com/topic/189150-validate-input-value-against-calculated-value/#findComment-998605 Share on other sites More sharing options...
kool_samule Posted January 20, 2010 Author Share Posted January 20, 2010 Hi, should have been a bit clearer I guess, $price_total=['calculation'] <input type='text' name='jobquote' id="count" class='price_warning' value="<?php echo number_format($price_total, 1, '.', '').'0'; ?>"/> So $price_total is the default value of the input, but it can be changed, and if changed, i need the 'admin override' option to appear.. . . Link to comment https://forums.phpfreaks.com/topic/189150-validate-input-value-against-calculated-value/#findComment-998612 Share on other sites More sharing options...
Buddski Posted January 20, 2010 Share Posted January 20, 2010 Ok you need to create a hidden field called like "original_job_quote" and submit that with the form.. So when the form is submitted get it to check the 2 values; echo '<input type="input" name="jobquote" value="'.$price_total.'" /> <input type="hidden" name="original_jobquote" value="'.$price_total.'" />'; // On Post // if ($_POST['original_jobquote'] != $_POST['jobquote']) { // They dont match } Link to comment https://forums.phpfreaks.com/topic/189150-validate-input-value-against-calculated-value/#findComment-998637 Share on other sites More sharing options...
kool_samule Posted January 21, 2010 Author Share Posted January 21, 2010 Got a bit further: <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? Link to comment https://forums.phpfreaks.com/topic/189150-validate-input-value-against-calculated-value/#findComment-999268 Share on other sites More sharing options...
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