On Thu, Mar 5, 2009 at 10:30 AM, Terion Miller <webdev.ter...@gmail.com>wrote:
> Still having problems with getting this script to work, the first part of
> the query does now work since I used the suggested JOIN, so my results are
> there and I can echo them but now I can't seem to get them to display
> neatly
> somehow:
>
> --------------------CODE THUS
> FAR--------------------------------------------
> $query = "SELECT admin.UserName, admin.AdminID, workorders.WorkOrderID,
> workorders.CreatedDate, workorders.Location, workorders.WorkOrderName,
> workorders.FormName, workorders.STATUS, workorders.Notes, workorders.pod
> FROM admin LEFT JOIN workorders ON (admin.AdminID = workorders.AdminID)
> WHERE admin.Retail1 = 'yes'
> ";
>
> $result = mysql_query ($query) ;
> //$row = mysql_fetch_array($result);
> while ($row = mysql_fetch_row($result)){
> $admin_id = $row['AdminID'];
mysql_fetch_row() returns a numerical array (
http://ca2.php.net/manual/en/function.mysql-fetch-row.php), but then you are
trying to assign $admin_id using an associative array. Thus, you need to
either return your row as an associative array (
http://ca2.php.net/manual/en/function.mysql-fetch-assoc.php) or assign
$admin_id as a numerical array:
Method 1: Use a numerical array
$result = mysql_query($query);
while($row = mysql_fetch_row($result)) {
$admin_id = $row[1]; // since it's the 2nd item in your SELECT
...
}
OR
Method 2: Use an associative array
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)) { // returns result row as an
associative array
$admin_id = $row['AdminID'];
....
}
