Question

I am trying to create a select drop-down list using php. Every time i try, i get an error.
Here is my code:

The function:

function dropDown(){
    $options="<select>"; 
    $connect = mysql_connect('localhost','id','pass') or die ("couldn't   connect!").mysql_error; 

    mysql_select_db('db') or die('could not connect to database!');

    $sql="SELECT * FROM DESC"; 
    $result=mysql_query($sql); 

    while ($row=mysql_fetch_array($result)) { // this is line 60
        $name=$row["name"]; 
    
        $options.="<option value=\"$name\">".$name."</option>"; 
    } 

    $options.= "</SELECT>";
    return "$options";
}

and then i just call it in my code

<?php  
include ('includes/functions.php');

// ...

$list = dropDown();
echo "$list";

// ...

?>

The error I get is:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/garagenj/public_html/dispatch/includes/functions.php on line 60

Answer 1

If your table name is DESC then that’s your problem. It’s all about namespacing. You can’t even have a field named desc, I’ve tried. It errors everytime