MYSQL: Insert variables into variable name table

Question

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`");

Share your few more lines of code to understand the problem. — Suresh, Oct 06 '18 at 07:23
Do you have error reporting set - https://stackoverflow.com/questions/22662488/how-to-get-mysqli-error-information-in-different-environments — Nigel Ren, Oct 06 '18 at 07:24
@NigelRen, no, I will set it now and I will try again. Thank you — Daniel S, Oct 06 '18 at 07:26
@NigelRen, where can I see mysqli errors from mysqli_report, if there is one? In console nothing shows up. — Daniel S, Oct 06 '18 at 07:32
@Suresh I tried now the table with hardcoded name, without a variable and it still does not work. Can it be due to other variables values? — Daniel S, Oct 06 '18 at 07:45
@DanielS: May be there is mismatch of column and values. Please share your table structure. — Suresh, Oct 06 '18 at 07:47
@DanielS : please check answer. I have made change in insert query with mentioning column name in it. — Suresh, Oct 06 '18 at 07:56

score 1 · Accepted Answer · answered Oct 06 '18 at 07:54

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")));
}

?>

@DanielS : Most welcome... Remember to use column name in insert query if you are skipping any column name from insert. Enjoy Coding ! :) — Suresh, Oct 06 '18 at 07:58

MYSQL: Insert variables into variable name table

1 Answers1