Unknown column 'E5V7D24M10Y2015' in 'where clause' - php

Question

I have the code below to update (more 'add to') a row in a database. I have read a few very similar posts, but still can't see where I'm going wrong...
I am getting the error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''EventID' = '2' WHERE GameID = 'E2V1D24M10Y2015' AND PlayerID = '55'' at line 1
In this case, 'E5V7D24M10Y2015' is the value of $GameID. I am looking for a column called GameID where the value is E5V7D24M10Y2015, not a column of that name. Please tell me why my $sql is looking for a column named after the value it is looking for.
Each time the function runs the count($Runners) will be different and the values in each variable will be different. That is why I have the SQL in a loop.
if ($formtype == "gameresults"){
$Runners = $_POST['runners'];
$event = $_POST['event'];
$eid = $_POST['eid'];
$vid = $_POST['vid'];
$GameID = $_POST['GameID'];
$date = $_POST['date'];
$season = $_POST['season'];
$region = $_POST['region'];
$notes = $_POST['notes'];
$kev = "kev#email.com";
$email = $_POST['manager'];
$notes = wordwrap($notes,70);
$subject = ("Results for " . $event);
$tix = "";
$cashs = "";
for ($x = 1; $x < ($Runners + 1); $x++){
$ID = $_POST['ID' . $x];
$Name = $_POST['Name' .$x];
$Place = $_POST['Place'.$x];
$Points = $_POST['Points'.$x];
$Cash = $_POST['Cash'.$x];
$Ticket = $_POST['Ticket'.$x];
$vn = $_POST['vn'];
$buyin = $_POST['buyin'];
$data = array($eid,$vid,$region,$buyin,$GameID,$date,$season,$ID,$Name,$Place,$Points,$Ticket,$Cash,$Runners);
$fields = array('EventID','VenueID','Region','Buyin','GameID','Date','Season','PlayerID','Name','Position','Points','Ticket?','Cash','Runners');
for ($x = 0; $x < (count($data) - 1); $x++){
$sql = "UPDATE results SET '$fields[$x]' = '$data[$x]' WHERE GameID = '$GameID' AND PlayerID = '$ID'";
$res = mysqli_query($dbcon, $sql) or die("Update failed. <br>" . mysqli_error($dbcon));
}

Answer 1

You could modify your code to be more clear. I did example but I did not map all the fields so you need to adjust it.
$eid = 4;
$vid = 5;
$region = 'aa';
$buyin = 'asd';
$gameID = 5;
$date = 5;
$playerID = 10;
$fields = array(
'EventID' => $eid,
'VenueID' => $vid,
'Region' => $region,
'Buyin' => $buyin,
'GameID' => $gameID,
'PlayerID'=> $playerID,
'Date' => $date,
);
$numItems = count($fields);
$query = "UPDATE `results` SET ";
$i = 0;
foreach($fields as $name => $value) {
++$i;
if($name == 'GameID' || $name == 'PlayerID') {
continue;
}
$query .= sprintf(" `%s` = '%s'%s ", $name, $value, ($i === $numItems ? "": ","));
}
$query .= sprintf(" WHERE `GameID` = '%d' AND `PlayerID` = '%d'", $fields['GameID'], $fields['PlayerID']);
echo $query;
Fields can be maped via $key => $value and then used in foreach loop. Also using sprintf() makes is more clear to read. However, the best option would be using prepared statements.
Also you don't need to make multiple UPDATE queries, just SET more parameters in one query.

Answer 2

$data = array($eid,$vid,$region,$buyin,$GameID,$date,$season,$ID,$Name,$Place,$Points,$Ticket,$Cash,$Runners);
$fields = array('EventID','VenueID','Region','Buyin','GameID','Date','Season','PlayerID','Name','Position','Points','Ticket?','Cash','Runners');
for ($x = 0; $x < (count($data) - 1); $x++){
$sql = "UPDATE results SET " . $fields[$x] . " = '$data[$x]' WHERE results.GameID = $GameID AND results.PlayerID = $ID";
$res = mysqli_query($dbcon, $sql) or die("Update failed. <br>" . mysqli_error($dbcon));
}
Try this by specifying alias.