On Wed, Oct 28, 2009 at 4:21 PM, Allen McCabe <allenmccabe@xxxxxxxxx> wrote:
> Hey everyone, I have an issue.
>
> I need my (employee) users to be able to insert shows into the our MySQL
> database and simultaneously upload an image file (and store the path in the
> table).
>
> I have accomplished this with a product-based system (adding products and
> uploading images of the product), and accomplished what I needed because
> the
> product name was unique; I used the following statements:
>
> $prodName = $_POST['prodName'];
> $prodDesc = $_POST['prodDesc'];
> $prodPrice = $_POST['prodPrice'];
>
> $query2 = "INSERT INTO product (pID, pName) VALUES (NULL, '$prodName');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query = "SELECT pID FROM product WHERE pName = '$prodName';";
> $result = mysql_query($query) or die(mysql_error());
> $row = mysql_fetch_array($result) or die (mysql_error());
>
> $prodID = $row['pID'];
>
>
> I had to select the new product to get the product id to use in the new
> unique image name.
>
> The problem I am facing now, is that with the shows that my users add will
> have multitple show times; this means non-unique titles. In fact, the only
> unique identifier is the show id. How can I insert something (leaving the
> show_id field NULL so that it is auto-assigned the next ID number), and
> then
> immediately select it?
>
> PHP doesn't seem to be able to immediately select something it has just
> inserted, perhaps it needs time to process the database update.
>
> Here is the code I have now (which does not work):
>
> $query2 = "INSERT INTO afy_show (show_id, show_title, show_day_w,
> show_month, show_day_m, show_year, show_time, show_price, show_description,
> show_comments_1, show_seats_reqd) VALUES (NULL, '{$show_title}',
> '{$show_day_w}', '{$show_month}', '{$show_day_m}', '{$show_year}',
> '{$show_time}', '{$show_price}', '{$show_description}',
> '{$show_comments_1}', '{$show_seats_reqd}');";
> $result2 = mysql_query($query2) or die(mysql_error());
>
> $query3 = "SELECT * FROM afy_show WHERE *show_id = '$id'*;";
> $result3 = mysql_query($query3) or die('Record cannot be located!' .
> mysql_error());
> $row3 = mysql_fetch_array($result3);
> $show_id = $row3['show_id'];
>
> How do I select the item I just inserted to obtain the ID number??
>
mysql_insert_id <http://ar.php.net/manual/en/function.mysql-insert-id.php>
mysqli->insert_id <http://ar.php.net/manual/en/mysqli.insert-id.php>
--
Martin Scotta
