我知道这是一个"众所周知"的问题,但我无法解决它。下面的MySQL查询(在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());
我试了很多不同的引语,但我都做不到。有人能把我送到正确的方向吗?
此外;
$column = 'prijs_max';
$vav = $INFO[$count+1]; // returning a number
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 ++;
}
尝试
"UPDATE $gameID SET `".$column."`='$vav' WHERE drank='$drank'";
"UPDATE $gameID SET $column ='$vav' WHERE drank='$drank'";