bytesize

Yes! The user pays to submit the photo. When they return from PayPal, some data is posted to the form. index.php case "create": $id_num = mysql_real_escape_string($_POST['txn_id']); $rs_duplicates = mysql_query("select id from photos where paypal_id='$id_num'"); $duplicates = mysql_num_rows($rs_duplicates); if ($duplicates > 0) { header("Location: index.php"); exit(); } if(is_jpeg($_FILES['photo']['type']['file']) and is_valid_file_size($_FILES['photo']['size']['file']) and is_uploaded_file($_FILES['photo']['tmp_name']['file']) and is_minimum_width_height($_FILES['photo']['tmp_name']['file']) and is_maximum_width_height($_FILES['photo']['tmp_name']['file']) ) { if(is_fields_filled_out($_POST['photo'])) { $_POST['photo']['theme_id'] = $this_weeks_theme['id']; $photo_id = create_photo($_POST['photo']); copy($_FILES['photo']['tmp_name']['file'], './photos/'.$photo_id.'.jpg'); $notice = "Successfuly created photo."; } else { $warning = "Click back and fill in all fields."; } } else { $warning = "Click back, error uploading file. Please make sure your file is a jpeg or gif and less than 1MB in size."; } break; } form.php <?php if(!empty($_POST)): ?> <p>Hello, <b><?php echo $_POST['first_name']; ?> <?php echo $_POST['last_name']; ?></b>, thank you for your order. Please check your email <b><?php echo $_POST['payer_email']; ?></b> for your receipt.</p> <h1>Submit Your Photo</h1> <p>Fill in the form and your Photo will be added to the Album of this web site!</p> <form action="index.php?view=create" method="post" enctype="multipart/form-data"> <fieldset> <div> <label><b>Web Address:</b></label> <input name="photo[discuss_url]" size="40" type="text" value="http://" class="textfield" /> <br/> <label>This is the link to your web site. Please prefix with: http://</label> </div> <div> <label><b>Attach Photo</b></label> <input name="photo[file]" type="file" class="textfield" /><br/> </div> <input name="photo[paypal_id]" size="40" type="hidden" value="<?php echo $_POST['txn_id']; ?>" /> <input type="submit" name="submit" value="Upload Photo" onSubmit='return clearForm(form_new)' /> </fieldset> </form> <?php endif; ?> create.php <?php header("Location: index.php"); ?>

This code is in form.php. The POST txn_id is added to the paypal_id column in the photos table when the form is submitted. <input name="photo[paypal_id]" size="40" type="hidden" value="<?php echo $_POST['txn_id']; ?> I use this code in index.php but there is no POST to this page. So it doesn't work! $id_num = mysql_real_escape_string($_POST['txn_id']); $rs_duplicates = mysql_query("select id from photos where paypal_id = '$id_num'"); $duplicates = mysql_num_rows($rs_duplicates); if ($duplicates > 0) { header("Location: index.php"); exit(); } I would like to check the column paypal_id for the insertion of txn_id that came from the form. If there's a match then exit. This would stop the user from clicking back and uploading the form and image again. Can someone help me with this?

It works! Thank you

Thanks for the quick reply. I tried your code. It's not working. It's probably me! I will continue to work with your code and report back later. Thanks.

This code is in form.php. The id_number is added to customer_id column in the photos table when the form is submitted. The id_number is automaticaly generated. <input name="photo[customer_id]" size="40" type="hidden" value="<?php echo $_POST['id_number']; ?>" /> Let's say 4P2792Q893421 was posted into the database. This code is in index.php and checks the database customer _id column for a match from id_number. If found, exit;, if not found, add the values to the database. As you can see, I manualy inserted the id_number into the code. It's checking the column customer_id and finds a match which in turn exits to index.php. $same_number= mysql_query("select id from photos where customer_id= '4P2792Q893421'"); $duplicate = mysql_num_rows($same_number); if ($duplicate > 0) { header("Location: index.php"); exit(); } I tried to POST the id_number in this example and it doesn't work. What's wrong? $same_number= mysql_query("select id from photos where customer_id= ''$_POST[id_number]'"); $duplicate = mysql_num_rows($same_number); if ($duplicate > 0) { header("Location: index.php"); exit(); }