Sudden

Try changing: $SQL = "DELETE FROM users WHERE ID = $id"; To: $SQL = "DELETE FROM users WHERE ID = ".$id; Not sure if thats the problem tho.

Ok...I was thinking storing info in sessions, and then testing it against them later. Which wont work. But your idea seems to be plausable.

http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_str-to-date Might help you out.

Use the switch function to test what their selection was and display the proper info.

Sessions go away after the browser is closed. So it probably wont help.

<img src= <?=$image?> /> should probly read <img src= <?php $image ?> />

If the first file contains PHP code it must be saved as .php

NP man Glad to be helpful

$ipaddy = $_SERVER['REMOTE_ADDR']; $query = mysql_query("SELECT ip FROM table WHERE ip='$ipaddy'"); Is correct syntax. Tested it. Just remember your gonna want to test their IP probably before you insert their info into your DB. Just a heads up.

NP you just have to have a statment that checks to see if the first form was submitted, and then use a 2nd form to make sure all the data is validated then you can send it off to the next site. It gets easier the more you use it.

Well...I'm not exactly sure what your aiming for, but i'm sure the following link can help you out a little: http://www.homeandlearn.co.uk/php/php4p11.html

You can intergrate the validation page w/ the site in which they input their information. <?php if(!isset($_POST['validate']){ //checks to see if they have tryed to validate their info ?> //form goes here with action = "<?php $_SERVER['PHP_SELF'] ?>" //fill out info //click validate button returning a value in $_POST['validate'] button HTML below <input type="submit" name="validate" value="Validate"> //website will reload w/ all values that were posted in the form, but the form will not be shown because of the if(!isset($_POST['validate'])) statement. <?php }else{ //all validation code should go here //another form could be displayed to give another submit button after all information checks out ?> <input type="submit" name="submit" value="Submit"> //This form should have the action below so it will post all information in the $_POST[] on the 2nd page <form action="_gdForm/webformmailer.asp" method="post"> <?php } ?> I hope this helps. Also there may be an easier way to do this.

Set it to insert the data into the correct column then give it the value the user input. Obviously change vars as needed. And make sure your method for submit is post if you plan on using $_POST[] to assign vars $input = $_POST['input']; $q = "INSERT INTO db_name (id) VALUES ('$input')"; mysql_query($q);

When you hit the submit button it will send all $_POST[] vars to the site specified in: <form action="_gdForm/webformmailer.asp" method="post"> From there you would just have to call the var Ex. $email = $_POST['email']; Sub email with what the value of your input box was on previous page. Then just set $email in the email link. All information reguarding $email, and $_POST['email'] should not be able to be viewed with source code. Hope this helps

Also the function: htmlspecialchars(); might help him handle the ' and ". It reads predefined php characters and turns them into HTML entities.