weird error - when pulling data into <select> (drop down)

[phence]

I am having trouble getting category tree stuctures from a database into <select> box (drop down menu).

with out the <select> code, everything works perfect, but when i try to load it into a <select> box nothing really shows up... but if i look at the "view source" i see this...

    <td width="29%"><font size="2" face="Tahoma">Select Category</font></td>
    <td width="32%"><SELECT name="categories">          

Error 0 : 

here is my php code...

<SELECT name="categories">          

<?php

// category functions that process category trees
function categories ($id, $depth) {
    if ($id == 'NULL')
        $query = "SELECT * FROM categories WHERE Parent IS NULL";
    else
        $query = "SELECT * FROM categories WHERE Parent = '" . $id . "'";

$result = mysql_query($query);
if (mysql_num_rows($result) < 1)
    return;
while($row = mysql_fetch_array($result)) {
    categories2 ($row, $depth);        
}
}
// function that puts everthing into a <select> or just prints it
function categories2 ($row, $depth) {
	if (($result = @ mysql_query ($query, $connection))){
    	while ($category = mysql_fetch_array($result)){
?>
<option name="<?php echo $category['ID']; ?>">
<?php 
	echo($depth . $row['Name']); ?></option><?php
    $depth = $depth . $row['Name'] . " > ";
    categories ($row['ID'], $depth); 
	// contiue if statement
		}
	}         

	else
        die("Error " . mysql_errno() . " : " . mysql_error());
}

// db data
$hostName = 'localhost';
$username = 'usrname';
$password = 'pw';
$databaseName = 'dbname';

//connect to DB
      if (!isset($connection))
      { 
          if (!($connection = @ mysql_connect($hostName, $username, $password)))
             showerror();

      if (!mysql_select_db($databaseName, $connection))
         showerror();

      $open = true;
  }

// Run the Functions
categories("NULL", "");
?>
</select>

does anyone know why this isn't working??
I would really appreciate your help

[Weedpacket]

Well, categories2() has no idea what $query and $connection are ... you wanna perhaps pass them as arguments, or declare them global? While you're there - what's $row supposed to be? You're using it as an array but passing it a string...

[phence]

Hey, Thanks for the reply...

Here is my fixed code.. still giving me that same error in the HTML source code.

<SELECT name="categories">
<?php
// category functions
function categories ($id, $depth) {
	if ($id == 'NULL')
        $query = "SELECT * FROM categories WHERE Parent IS NULL";
    else
        $query = "SELECT * FROM categories WHERE Parent = '" . $id . "'";

$result = mysql_query($query);
if (mysql_num_rows($result) < 1)
    return;
while($row = mysql_fetch_array($result)) {
    $ID = $row['ID']; 
	$Name = $row['Name']; 
	categories2 ($ID, $Name, $depth, $query, $connection);        
}
}
// function that puts everthing into a <select> or just prints it
function categories2 ($ID, $Name, $depth, $query, $connection) {
	if (($result = @ mysql_query ($query, $connection))){
    	while ($category = mysql_fetch_array($result)){
?>
<option name="<?php echo $category['ID']; ?>"> 
<?php 
	echo($depth . $Name); 
?>
</option>
<?php
    $depth = $depth . $Name . " > ";
    categories ($ID, $depth); 
	// contiue if statement
		}
	}         

	else
        die("Error " . mysql_errno() . " : " . mysql_error());
}

// db data
$hostName = 'localhost';
$username = 'name';
$password = 'pw';
$databaseName = 'datname';

//connect to DB
  if (!isset($connection))
  { 
      if (!($connection = @ mysql_connect($hostName, $username, $password)))
         showerror();

      if (!mysql_select_db($databaseName, $connection))
         showerror();

      $open = true;
  }

// Run the Functions
categories("NULL", "");
?>
      </select> 

thanks!

[phence]

anyone know why this isn't working? 🙁

[Weedpacket]

From what I can make out, you're dying with an error even when there's nothing wrong except that your query doesn't return any results. It's not MySQL's fault if nothing matches the query - which is why the error number is 0 and the error message is blank: as far as MySQL is concerned, there was no error. But you die() anyway.

if (($result = @ mysql_query ($query, $connection))){...}else{die(...)}

(PS. It's often a good idea to leave @ off when you're developing. You don't know what you're missing, otherwise.)

Me, I'd go more like

$result = mysql_query(...) or die(...);
$result_count = mysql_num_rows($result);
if($result_count) // or !=0 or  >0 ...
{ for($i=0; $i<$result_count; $i++)
  { $category = mysql_fetch_array(...);
    ...
  }
}
else
{ ... Gracefully handle the case when no result were returned ...
}

[phence]

hey thanks alot weed!
at lease now i'm getting a reasonable error:
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/www/user/test/ya/creat_art.php</b> on line 35
Error 0 :

this is line 35:

	$result = mysql_query($query, $connection) or die("Error " . mysql_errno() . " : " . mysql_error());

thanks for the help! what's this error ?? how can i fix it?

[Weedpacket]

I notice that categories() has no idea what $connection is either.