I am trying to make a simple database delete script.

Im fairly new to php and mysql.

I need to access 3 tables in the database to delete all the relevany information for one user.

The articles, penname and users tables.

I adapted the code from a search script I put together.

The following code seems to work if the user has a penname and some articles, however doesnt seem to work if the user has no pennames or articles.

I probably need some if statements.

Thanks for any help.

This is the code:

<h2>Search</h2> 
<form name="search" method="post" action="<?=$PHP_SELF?>">
Search for: <input type="text" name="find" /> in 
<Select NAME="field">
<Option VALUE="email">Email Address</option>
</Select>
<input type="hidden" name="searching" value="yes" />
<input type="submit" name="search" value="Search" />
</form>

<?
include('setup.php'); 
//This is only displayed if they have submitted the form 
if ($searching =="yes") 
{ 
echo "<h2>Results</h2><p>"; 
//If they did not enter a search term we give them an error 
if ($find == "") 
{ 
echo "<p>You forgot to enter a search term"; 
exit; 
} 
// Otherwise we connect to our Database 
@mysql_connect($dbhost,$dbuser,$dbpasswd)or die("Connect Error: ".mysql_error());
mysql_select_db($dbname) or die( "Unable to select database");
// We preform a bit of filtering 
$find = strtoupper($find); 
$find = strip_tags($find); 
$find = trim ($find); 
//Now we search for our search term, in the field the user specified 
// Construct our join query
$data = mysql_query("DELETE articles, penname, users ".
 "FROM articles, penname, users ".
 "WHERE articles.penname_id = penname.user_id AND penname.user_id = users.user_id AND upper($field) ='$find'"); 
  
 
if ($anymatches == 0) 
{ 
echo "Sorry, but we can not find an entry to match your query<br><br>"; 
}  
//And we remind them what they searched for 
echo "<b>The Records You Searched For:</b> " .$find; 
} 
?>

It's not working because the query is requiring the data to be present in each table.

You're going to need separate queries for each table.