Error: Not a valid sql result

[mikec87]

I am trying to populate a dropdown, but cannot get the sql to return what I want. Can someone please help me with this error that I am getting in case 1? I am not using case 2 or the default case yet.

"<br />
<b>Warning</b>: mysql_fetch_arr ay(): supplied argument is not a valid MySQL result resource in <b>/home/user21/public_html/mp3/enterjob_2.php</b> on line <b>23</b><br />
<br />
<b>Notice</b>: Undefined variable: temp in <b>/home/user21/public_html/mp3/enterjob_2.php</b> on line <b>26</b><br />

"
Ultimately, I want to use the results to populate a dropdown, and am trying to get the data in a format that I can use in javascript, but keep getting the error. I am trying to make a name:value pair, so I can move over to the javascript side and manipulate the string and insert values into an existing dropdown. If there is a better way, please let me know. Thank you.

My php code is:
[php]
<?php
$host="localhos t"; // Host name
$username="mysq lusername"; // Mysql username
$password="mysq lpassword"; // Mysql password
$db_name="testd b"; // Database name

if($_GET['action'] == 'getOptions'){
// Connect to server and select databse.
mysql_connect(" $host", "$username" , "$password" )or die("cannot connect");
mysql_select_db ("$db_name") or die("cannot select DB");

switch($_GET['jobtype']){
case 0:

break;

case 1:
error_reporting (E_ALL);
ini_set('displa y_errors', True);

$sql="Select id, yardtype From YardTypes";

while ($row = mysql_fetch_arr ay($sql) )
$temp = $temp.$row[id].":".$row[yardtype].";";

echo ($temp);

break;

case 2: //Internet
break;

default: //User has selected all required information...s ave job
$jobtype=$_POST['jobtype'];
$jobdescription =$_POST['jobdescription '];

$sql="INSERT INTO $tbl_name (userid, jobtype, jobdescription) VALUES('$userid ', '$jobtype', '$jobdescriptio n')";

if (mysql_query($s ql))
header("locatio n:thankyou.php" );
else
echo "The server is temporarily unavailable. We apologize for the inconvenience." ;

break;
}
}
?>
[/php]

[pbmods]

Heya, Mike.

Before calling mysql_fetch_arr ay(), you must first execute your SQL query and generate a result resource by using mysql_query():

$sql="Select id, yardtype From YardTypes";
 
$res = mysql_query($sql);
 
if( empty($res) )
{
    echo $sql, '<br /><br />', mysql_error();
}
 
while( $row = mysql_fetch_assoc($res) )
{
    .
    .
    .
 

[mikec87]

Heya, Mike.

Before calling mysql_fetch_arr ay(), you must first execute your SQL query and generate a result resource by using mysql_query():

$sql="Select id, yardtype From YardTypes";
 
$res = mysql_query($sql);
 
if( empty($res) )
{
    echo $sql, '<br /><br />', mysql_error();
}
 
while( $row = mysql_fetch_assoc($res) )
{
    .
    .
    .
 

That worked great...thank you very much for the second set of eyes. I guess I have been working too much today.

[pbmods]

Heya, Mike.

Glad to hear you got it working! Good luck with your project, and if you ever need anything, well, you know... :)