Dynamic Select list using sql data

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

Answer 2

Either your table is using a reserved word (desc) or your query is all jacked up. Try this:

select * from table order by col desc

Be sure to replace table and col with your table name and column name.

Answer 3

SELECT * FROM DESC ??????

• From what table
• Or, if the table name is DESC. escape them with ticks

---

And You dont have start a new connection, every time you call the function. Put your connection part somewhere else

$connect = mysql_connect('localhost','id','pass') or die ("couldn't   connect!").mysql_error; 
mysql_select_db('db') or die('could not connect to database!');

function dropDown(){
    $options="<select>"; 
    $sql="SELECT * FROM `DESC`"; 
    $result=mysql_query($sql); 
    
    while ($row=mysql_fetch_array($result)) { 
    
        $name=$row["name"]; 
    
        $options.="<option value=\"$name\">".$name."</option>"; 
    } 
    
    $options.= "</SELECT>";
    return $options;

}

Answer 4

This is wrong:

$sql="SELECT * FROM DESC"; 

You forgot to add the table name to select from