Mysql not updating before next form

You could try adding a hidden form field with a value of the passcode, (which is terribly unsecure)...... Try this, echo out your $sql variable after the variable is created.

That will tell you what is being run on the database. Not sure why the ' wasn't around the passcode in my code above. Im trying to get my name from a form on page one to show up on the next page. Here is page 1: [PHP] Thanks for the responses back!

Note: If you have accessed this web page via a search engine, you should go back and start at the Beginning.

This tutorial is designed to be viewed and executed in sequence.

The process of updating or editing a record or row of a table makes use of the trusty mysql_query() function.

In the syntax shown below we see that the UPDATE procedure requires the SET and WHERE definitions to pinpoint the changes.

really need help on my beginner development :) Here's the case...

Here are the instructions to insert the data from page 2 into the same table as page 1 (test).

Im trying to get it to go into the same row by using the following code. The logic seems so simple but I just cant get it to work. The column names are passcode, name, address, city, state, zip and email.

you have: [PHP]$passcode = $_POST[passcode];[/PHP] which I think should be: [PHP]//note the single quotes inside the brackets $passcode =$_POST['passcode'];[/PHP] If that doesn't work, post back here and we'll see if we can figure it out.

Greg Hi, you might also want to consider changing your code - example from this : [PHP]$passcode = $_POST[passcode]; $name = $_POST['name']; mysql_query("INSERT INTO test (passcode, name, address, city, state, zip, email) VALUES('$passcode','$name','a','b','c','d','e') ") or die(mysql_error()); This page also has another form to fill out with the variables $address, $city, $state, $zip, $email.[/PHP] to [PHP]$passcode = $_POST['passcode']; $name = $_POST['name']; mysql_query("INSERT INTO test (passcode, name, address, city, state, zip, email) VALUES('".$passcode."','".$name."','a','b','c','d' ,'e') ") or die(mysql_error()); This page also has another form to fill out with the variables $address, $city, $state, $zip, $email.[/PHP] Regards Purple Sorry.

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