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how to define $id


herghost

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Hi all, I am having a ponder on how you would go about this:

 

Basically I have a page from where all the users in a database are shown, and you can access a page called userdetails.php by taking the id from the database ie userdetails.php?id=1

 

Once on this page I want to be able to add more details to various tables, so on this page I will have a various links for new forms eg.

 

add services

add prices

 

eg.

 

How would I make sure that in each form the id number matched the users id?

 

would i have to do something like addservices.php?id=$id ?

 

Cheers

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Thanks,

 

now getting Notice: Undefined index: idaddservicesdo.php on line 6

 

Might be easier if just show you what I have!

 

addservices.php

 

<form method="post" action="addservicesdo.php">
<?php
if (isset($_GET['id'])) {
  echo "<input type='hidden' name='id' value='{$_GET['id']}' />";
}
?>
  <p>hosting package
    <select name="hosting">
      
      <option value="">None</option>
      <option value="standard">Standard</option>
      <option value="professional">Professional</option>
      <option value="business">Business</option>
    </select>
    
    <br>
    
    domain: 
    <input type="text" name="domain">
    <br>
    
    Backup plan 
    <select name="backup">
      <option value="">none</option>
      <option value="500mb">500mb</option>
      <option value="1000mb">1000mb</option>
      <option value="5000mb">5000mb</option>
      <option value="1000mb">1000mb</option>
    </select>
     <input type="submit" value="Next">
  </p>
</form>

 

addservicesdo.php

 

<?php

include('../include/dbconfig.php');
//let's start the session
session_start();
$id  = ($_POST['id']);
$hosting = ($_POST['hosting']); 
$domains = ($_POST['domain']);
$backup = ($_POST['backup']);


mysql_select_db("jolly_accounts", $conn);
$sql="INSERT INTO services (id,hosting,domains,backup) VALUES ('$id', '$hosting', '$domains', '$backup')"; 
if (!mysql_query($sql,$conn)) {
die('Error: ' . mysql_error());
}
echo "services successfully added to user.";
mysql_close($conn);
?>

 

So much for 'this will be easy!'

 

Thanks guys

 

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