Use mysql_insert_id() instead of the select you´re using now. "Allen McCabe" <allenmccabe@xxxxxxxxx> escreveu na mensagem news:657acef20910281221y5ab6ab7t4882f4f00da2cd19@xxxxxxxxxxxxxxxxx > 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?? > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
