Re: Re: Problems with displaying results
| From: | Lex Braun | Date: | Thu, 05 Mar 2009 17:25:24 +0000 |
| Subject: | Re: Re: Problems with displaying results | ||
| References: | 1 2 3 4 5 | Groups: | php.db php.general |
| Request: | Send a blank email to php-db+get-46035@lists.php.net to get a copy of this message | ||
On Thu, Mar 5, 2009 at 10:30 AM, Terion Miller <webdev.terion@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'];
....
}