php - SQL 错误(更新 gegevens)::'1' 中的未知列 'field list'

Question

我知道这是一个“众所周知”的问题，但我无法让它发挥作用。以下 MySQL 查询(PHP 中)给了我这个错误

 $sqle = "UPDATE $gameID SET `$column` = `$vav` WHERE drank='$drank'";
 $resulte = mysql_query($sqle) or die('SQL Error (update gegevens):: '.mysql_error());

我尝试了很多不同的引用，但无法使其正常工作。有人可以带我走向正确的方向吗？

还有；

$column = 'prijs_max';
$vav = $INFO[$count+1];  // returning a number

<小时/>

编辑后的完整循环

$count = 0;
    foreach ($INFO as $value) {
        $column = "";
        if(strpos($value, '§') !== false) {
            $pieces = explode('§', $value);
            $drank = $pieces[0];
            $rang = $pieces[1];

            if ($rang == 'start') {
                $column = 'prijs_start';
            } elseif ($rang == 'min') {
                $column = 'prijs_min';
            } elseif ($rang == 'max') {
                $column = 'prijs_max';
            }

            if ($column == 'prijs_start') {
                $bidmaxquery = "SELECT drank FROM $gameID WHERE drank = '$drank'";
                $bidmax = mysql_query($bidmaxquery) or die('SQL Error (get drank) :: '.mysql_error());

                if (mysql_num_rows($bidmax) == 0) {
                    $vav = $INFO[$count+1];
                    $inc = $INFO[$count+7];
                    $sqld = "INSERT INTO $gameID (drank,$column,prijs_current,increment) VALUES ('$drank','$vav','$vav','$inc')";
                    $queryd = mysql_query($sqld) or die('SQL Error (insert eerste gegevens):: '.mysql_error());
                }
            } else {
                $vav = $INFO[$count+1];
                echo $vav;
                echo "<br>";
                $sqle = "UPDATE $gameID SET `".$column."`=$vav WHERE drank='$drank'";
                $resulte = mysql_query($sqle) or die('SQL Error (update gegevens):: '.mysql_error());
            }

        }
        $count ++;
    }

Answer 1

尝试

"UPDATE $gameID SET `".$column."`='$vav' WHERE drank='$drank'";