CODEWITHSUNDEEP
phpmysql
Daniel S
Daniel S

Reputation: 49

MYSQL: Insert variables into variable name table

I am trying to insert PHP variables in a mysql table, where the table name is also a variable, using mysqli_query. I've tried multiple solutions from stackoverflow but it still does not work.

I try to do it like this, maybe I am missing something. Thank you in advance!

<?php
session_start();
@include_once "modules/connections/dbconn.php";

$value = $_POST['value'];
$playerid = $_SESSION["steamid"];
$playername = fetchinfo("name","users","steamid",$playerid);
$playeravatar = fetchinfo("avatar","users","steamid",$playerid);
$playercoins = fetchinfo("coins", "users","steamid",$playerid);

if($playercoins - $value < 0){
    die(json_encode(array('message' => 'ERROR', 'code' => "Not enough coins!")));
}

$game = fetchinfo("value","parameters","name","raffleRound");
$maxitems = fetchinfo("value","parameters","name","raffleMaxritems");
$items = fetchinfo("itemsnum","rafflegames","id",$game);

$itemname = "Coins";
$itemavatar = "images/creditcardicon.png";
$color = "D2D2D2";

$initialvalue = fetchinfo("value","rafflegames","id",$game);
$from = $initialvalue * 100;
$to = $from + $value * 100;

$tablename = 'rafflegame'.$game;

if($items < $maxitems){
    mysqli_query($GLOBALS["connect"], "UPDATE rafflegames SET `value`=`value`+$value, `itemsnum`=`itemsnum`+1 WHERE `id`=$game");
    mysqli_query($GLOBALS["connect"], "UPDATE users SET `coins`=`coins`-$value WHERE `steamid`=$playerid");
    mysqli_query($GLOBALS["connect"], "INSERT INTO `" . $tablename . "` VALUES ('".$playerid."', '".$playername."','".$itemname."','".$color."','".$value."','".$playeravatar."','".$itemavatar."','".$from."','".$to."')");
}
else {
    die(json_encode(array('message' => 'ERROR', 'code' => "Too many items in the current game")));
}

?>

The other two queries work just fine.

The table structure is this:

mysqli_query($GLOBALS['connect'],"CREATE TABLE `rafflegame$roundNumber` (
  `id` int(11) NOT NULL auto_increment,
  `userid` varchar(70) NOT NULL,
  `username` varchar(70) NOT NULL,
  `item` text,
  `color` text,
  `value` float,
  `avatar` varchar(512) NOT NULL,
  `image` text NOT NULL,
  `from` int NOT NULL,
  `to` int NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=1 DEFAULT CHARSET=utf8 AUTO_INCREMENT=1;");
mysqli_query($GLOBALS['connect'],"TRUNCATE TABLE `rafflegame$roundNumber`");

Upvotes: 1

Views: 73

Answers (1)

Suresh
Suresh

Reputation: 489

There is difference between table structure and insert column coumnt, When you want id column as auto incremented in that case column name should be included in insert query.

Please use the code as below:

<?php
session_start();
@include_once "modules/connections/dbconn.php";

$value = $_POST['value'];
$playerid = $_SESSION["steamid"];
$playername = fetchinfo("name","users","steamid",$playerid);
$playeravatar = fetchinfo("avatar","users","steamid",$playerid);
$playercoins = fetchinfo("coins", "users","steamid",$playerid);

if($playercoins - $value < 0){
    die(json_encode(array('message' => 'ERROR', 'code' => "Not enough coins!")));
}

$game = fetchinfo("value","parameters","name","raffleRound");
$maxitems = fetchinfo("value","parameters","name","raffleMaxritems");
$items = fetchinfo("itemsnum","rafflegames","id",$game);

$itemname = "Coins";
$itemavatar = "images/creditcardicon.png";
$color = "D2D2D2";

$initialvalue = fetchinfo("value","rafflegames","id",$game);
$from = $initialvalue * 100;
$to = $from + $value * 100;

$tablename = 'rafflegame'.$game;

if($items < $maxitems){
    mysqli_query($GLOBALS["connect"], "UPDATE rafflegames SET `value`=`value`+$value, `itemsnum`=`itemsnum`+1 WHERE `id`=$game");
    mysqli_query($GLOBALS["connect"], "UPDATE users SET `coins`=`coins`-$value WHERE `steamid`=$playerid");
    mysqli_query($GLOBALS["connect"], "INSERT INTO `" . $tablename . "`(`userid`,`username`,`item`,`color`,`value`,`avatar`,`image`,`from`,`to`) VALUES ('".$playerid."', '".$playername."','".$itemname."','".$color."','".$value."','".$playeravatar."','".$itemavatar."','".$from."','".$to."')");
}
else {
    die(json_encode(array('message' => 'ERROR', 'code' => "Too many items in the current game")));
}

?>

Upvotes: 1

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