You all are great! Thank you so much.
On Wed, Oct 28, 2009 at 12:27 PM, Martin Scotta <martinsco...@gmail.com>wrote:
>
>
> On Wed, Oct 28, 2009 at 4:21 PM, Allen McCabe <allenmcc...@gmail.com>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
>
