joebudden Posted January 15, 2007 Share Posted January 15, 2007 Hi guys im creating a game in php and have a page which generates a game id[code]<?php//start the games sessionsession_start(); //connect to databaseinclude("connect.php");//generate random game id$gameid = rand(0,1000000);//if form is submitted add details to the tablerequire("start.view.php");?>[/code] [code]<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /><title>PHP Football Manager</title><link rel=stylesheet href="style.css" type="text/css"></head><body><table cellpadding="0" cellspacing="0"><tr><td><div class="title"></div></td></tr><tr><td><br /><a href='players.php?gameid=<?php echo $gameid;?>'>Start</a></td></tr></table></body></html>[/code]When the user clicks the start link, the game id generated is sent to the players page.This is where my problem is..If i use the code... [code]<?php$gameid = $_GET['gameid'];echo $gameid . '<br />';?>[/code]It echoes out the game id ok.But when i try to insert it into the database it says undefined index and variable..???Here is my code for that page[code]<?php//start the games sessionsession_start(); //connect to databaseinclude("connect.php");//query to determine wether player table is empty$ap_query = "select id,name,surname,ability,fitness,skill,position,prefmove,description,teamtalk from player order by surname";$allplayers = mysql_query($ap_query);//query to select total of players $tp_query = "select count(id) as total from player"; $total_players = mysql_query($tp_query);$a = mysql_fetch_array($total_players, MYSQL_ASSOC);//var containing total of players$total = $a['total'];//vars to hold player details $id = "id";$name = "name";$surname = "surname";$ability = "ability";$fitness = "fitness";$skill = "skill";$position = "position";$prefmove = "prefmove";$description = "description";$teamtalk = "teamtalk";//if form is submitted add details to the tableif(isset($_POST["add_players"]) && isset($_POST["gamename"])){//vars to hold insert data$gameid = $_GET['gameid']; //THIS IS THE GAMEID VARIABLE$players = $_POST["add_players"];$gamename = $_POST["gamename"];$test = count($players);echo $gamename . '<br />';if(($test > 5 ) || ($test < 5)){echo 'You have not selected 5 players <br />';}else{ foreach ($_POST["add_players"] as $players) { //insert query $addsql = "insert into game (id,gameid,gamename) values ('$players','$gameid','$gamename')"; mysql_query($addsql) or die (mysql_error()); }//query display players selected$testquery = "select name,surname from player,game where player.id = game.id and gameid = '$gameid'";$test2 = mysql_query($testquery);echo '';if (mysql_num_rows($test2) !== 0){while($row = mysql_fetch_assoc($test2)){echo $row['name'] . ' ';echo $row['surname'] . '<br />';}} }}else{echo 'No players chosen';}if (isset($_POST['delete'])){ $deletesql = "delete from game where gameid = '$gameid'"; //$deletesql = "delete from game"; mysql_query($deletesql) or die (mysql_error());}require("players.view.php");?>[/code]I know its something simple so if anyone has any ideas please let me knowcheers guys Link to comment https://forums.phpfreaks.com/topic/34276-inserting-get_-variable-into-database/ Share on other sites More sharing options...
Philip Posted January 15, 2007 Share Posted January 15, 2007 Try this:[code]<?php//start the games sessionsession_start(); //connect to databaseinclude("connect.php");//query to determine wether player table is empty$ap_query = "select id,name,surname,ability,fitness,skill,position,prefmove,description,teamtalk from player order by surname";$allplayers = mysql_query($ap_query);//query to select total of players $tp_query = "select count(id) as total from player"; $total_players = mysql_query($tp_query);$a = mysql_fetch_array($total_players, MYSQL_ASSOC);//var containing total of players$total = $a['total'];//vars to hold player details $id = "id";$name = "name";$surname = "surname";$ability = "ability";$fitness = "fitness";$skill = "skill";$position = "position";$prefmove = "prefmove";$description = "description";$teamtalk = "teamtalk";//if form is submitted add details to the tableif(isset($_POST["add_players"]) && isset($_POST["gamename"])){//vars to hold insert data$gameid = $_GET['gameid']; //THIS IS THE GAMEID VARIABLE$players = $_POST["add_players"];$gamename = $_POST["gamename"];$test = count($players);echo $gamename . '<br />';if(($test > 5 ) || ($test < 5)){echo 'You have not selected 5 players <br />';}else{ foreach ($_POST["add_players"] as $players) { //insert query $addsql = "insert into `game` (`id`, `gameid`, `gamename`) values ('".$players."','".$gameid."','".$gamename."')"; mysql_query($addsql) or die (mysql_error()); }//query display players selected$testquery = "select `name`,`surname` from `player`,`game` where player.id = game.id and gameid = '".$gameid."'";$test2 = mysql_query($testquery);echo '';if (mysql_num_rows($test2) !== 0){while($row = mysql_fetch_assoc($test2)){echo $row['name'] . ' ';echo $row['surname'] . '<br />';}} }}else{echo 'No players chosen';}if (isset($_POST['delete'])){ $deletesql = "delete from `game` where `gameid` = '".$gameid."'"; //$deletesql = "delete from game"; mysql_query($deletesql) or die (mysql_error());}require("players.view.php");?>[/code] Link to comment https://forums.phpfreaks.com/topic/34276-inserting-get_-variable-into-database/#findComment-161203 Share on other sites More sharing options...
joebudden Posted January 15, 2007 Author Share Posted January 15, 2007 Nope still get the same error.Anyone else? Link to comment https://forums.phpfreaks.com/topic/34276-inserting-get_-variable-into-database/#findComment-161219 Share on other sites More sharing options...
Hypnos Posted January 15, 2007 Share Posted January 15, 2007 Echo $addsql, and paste the insert query here. Link to comment https://forums.phpfreaks.com/topic/34276-inserting-get_-variable-into-database/#findComment-161249 Share on other sites More sharing options...
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