DB ERROR: SQL syntax; [closed]

2

I'm trying to create a temporary table with football team data, but I get this error:

  

DB ERROR: You have an error in your SQL syntax; check the manual that   correspond to your MySQL server version for the right syntax to use   near) and t.situacaojogador = 'active'

Can anyone help me understand what's wrong? Follow my code.

$sql = "select codtime from jogador where codjogador=" . $_SESSION['codjogador'] . ";";
$rst = my_query($connR, $sql);
if(count($rst) > 0) {
    $codtimejogadorlogado = $rst[0]['codtime'];
} else {
    $codtimejogadorlogado = -1;
}

if($codtimejogadorlogado == '') {
    $codtimejogadorlogado = '-1';
}


$sql = "create temporary table tmptime as
    (select t.codtime, t.nome, j.codjogador
    from time t
    left join jogador j on j.codtime=t.codtime
    where j.codtime in ($codtimejogadorlogado) and t.situacaojogador = 'ativo');";
my_execute($connR, $sql);

Solved, I put another if to check if the codtime was empty. Updated code.

    
asked by anonymous 26.09.2017 / 15:13

0 answers