franknu Posted May 8, 2007 Share Posted May 8, 2007 I have this code that when the user provide his email address he will participate in a raffle, my problem is that i want to select the id and the name but my selection is not workin here is my display 0 .Thank You. you are participarin to win a Brand new Laptop Here is your Lucky Number here is my code $sql="INSERT INTO `list` (`name`,`email`,`state`,`city`, `residential`,`commercial`) VALUES ('".$name."','".$email."','".$state."','".$city."','".$residential."','".$commercial."')"; $result = mysql_query($sql); echo mysql_error(); if($result) { $query = "SELECT * FROM `list` WHERE `email_id`= '$email_id' "; $result = mysql_query($query) or die (mysql_error()); $row = mysql_fetch_assoc($result); echo mysql_affected_rows()." .Thank You. $name<br> you are participarin to win a Brand new Laptop<br> Here is your Lucky Number <br>"; echo "$email_id "; } Link to comment https://forums.phpfreaks.com/topic/50549-selecting/ Share on other sites More sharing options...
per1os Posted May 8, 2007 Share Posted May 8, 2007 The reason being is $email_id is not set anywhere. You may want to look into www.php.net/mysql_insert_id Link to comment https://forums.phpfreaks.com/topic/50549-selecting/#findComment-248381 Share on other sites More sharing options...
franknu Posted May 8, 2007 Author Share Posted May 8, 2007 so there is not way of doing it the way i was doing it Link to comment https://forums.phpfreaks.com/topic/50549-selecting/#findComment-248409 Share on other sites More sharing options...
per1os Posted May 8, 2007 Share Posted May 8, 2007 Not unless you set $email_id to have some type of a value, given the code you posted above $email_id is empty/null/nothing. You never instantiated/defined it. Link to comment https://forums.phpfreaks.com/topic/50549-selecting/#findComment-248415 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.