[SOLVED] Help with option box please!

[littlevisuals]

Hi all

 

How do I create an option dropdown using data from mysql?

 

I have 2 fields (artist_firstname, artist_lastname) I need merging together in this box (as well as ordering alphabetically).  I've tried this code and it doesn't work, but then again im not sure if im even selecting the tables correctly 

 

My database ARTISTS

 

tables in ARTISTS are "ARTISTS, CATORGORIES & GALLERY"

 

here is the code

 

<label for="artist">Artists</label><SELECT NAME="artist">

<?php
mysql_connect("localhost","root","password");
mysql_select_db("artists");
   
$q = "SELECT artist_firstname, artist_lastname FROM `artists` ORDER BY name ASC";
$r = mysql_query($q);
while($column = mysql_fetch_assoc($r)){
$artist_lastname = $column['id'];
$artist_lastname = $column['artist_lastname'];
echo '<OPTION VALUE="'.$artist_firstname.'">'.$artist_lastname.'</option>';
      }
?>

</SELECT>

 

How would I insert the data (artist_firstname + artist_lastname) into one field?

 

Sorry if its something stupid i've just started learning php/mysql! 

[AwptiK]

echo '<OPTION VALUE="'.$artist_firstname.'.$artist_lastname.'">'.$artist_firstname.' '.$artist_lastname.'</option>';

 

Ex:

<option value="BonJovi">Bon Jovi</option>

[Psycho]

<?php

mysql_connect("localhost","root","password");
mysql_select_db("artists");

$query = "SELECT id, CONCAT(artist_firstname, ' ', artist_lastname) as name
          FROM `artists` ORDER BY artist_lastname ASC";
$result = mysql_query($query);

$artist_options = '';
while($row = mysql_fetch_assoc($r))
{
    $artist_options .= "<option value=\"{$row['id']}\">{$row['name']}</option>\n";
}
?>
<label for="artist">Artists</label>
<SELECT NAME="artist">
<?php echo $artist_options; ?>
</SELECT>

[littlevisuals]

Hi thanks everyone for some solutions

 

with your code mjdamato i get

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in 

 

Sorry adding $result to the fetch did it 

 

Thankyou!

 

 

[Psycho]

Sorry adding $result to the fetch did it 

 

yeah, sorry, it is a pet peeve of mine when people use variable names that offer absolutely no value to identifying what the variable is. It may seem more efficient when initially writing the code, but it creates problems when having to edit the code later or if someone else needs to edit the code. So, I will sometimes go a little too far when editing peoples code on forums.