Re: [PHP-DB] Re: [PHP] Re: [PHP-DB] Re: Problems with displaying results

On Thu, Mar 5, 2009 at 10:30 AM, Terion Miller <[email protected]>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'];
    ....
}