Php Insert不会将itemID插入数据库表

好的，所以当有人从我的站点购买商品时，需要将该商品ID插入数据库，但由于某些原因，它没有这样做

$mysql_host = 'localhost'; //Leave at localhost  
$mysql_user = ''; //DB User  
$mysql_pass = ''; //DB Pass  
$mysql_db = ''; //DB Name  
$file = 'paypal.log'; //Paypal Log Name will be placed in the same location as your ipn.php file
$payer_email = $_REQUEST['payer_email'];
$ip = $_SERVER['REMOTE_ADDR'];

$time = date("F j, Y, g:i a");
$paylist = array("5.00" => 15017, "0.01" => 18830, "10.00" => 13840, "10.01" => 13842, "9.99" => 13729, "15.00" => 19111, "60.00" => 1046, "40.00" => 1050, "14.99" => 6199);


// connect db  

$db = mysql_connect($mysql_host, $mysql_user, $mysql_pass) or die(mysql_error());

$custom = stripslashes(ucwords(strtolower(trim($_REQUEST['custom']))));  
$receiver_email = $_REQUEST['receiver_email'];  
$payment_status = $_REQUEST['payment_status'];  
$mc_gross = $_REQUEST['mc_gross'];
mysql_select_db($mysql_db, $db);  
if ($_REQUEST['debug']){
print $payment_status . '\n';
print (isset($paylist[$mc_gross])) ? 1 : 0 . '\n';
print $receiver_email . '\n';
print $custom . '\n';
}
if ($payment_status == "Completed" && $receiver_email == "justincopp77@yahoo.com" && isset($paylist[$mc_gross])) {  


$query = "INSERT `status` SET `item` = '$itemid' ,`username` = '$custom'";  

$result2 = mysql_query($query) or die(mysql_error());  

$prem = mysql_fetch_array($result);  
$somecode = "'$time' '$custom' '$payer_email' '$mc_gross' '$ip'\r\n";

// figure out how much to give
$give = $paylist[$mc_gross];
$itemid = $prem['item'] + $give;  
// $points = mysql_query($prem)  
$qry2 = "SELECT `item` FROM `status` WHERE `username` = '$custom'";
// Log Paypal Transaction
$hak = fopen($file, "a");
fwrite($hak, $somecode);
fclose($hak);

$result2 = mysql_query($qry2) or die(mysql_error());
}
?>

您可以看到它没有将项目ID放入 有人知道为什么它不会将项目id插入数据库吗？ 我需要更改什么来修复它

---

提前感谢

您的sql语法错误

$query = "INSERT INTO `status` (item, username) VALUES ('$itemid' ,'$custom')";  

---

现在是抛弃这个不推荐的API的好时机，仔细阅读prepared语句您正在使用INSERT更新一行，这就是它不能按预期工作的原因。我认为它需要说INSERT INTO

status

，而不仅仅是INSERT'status'，您在赋值（第43行）之前使用$itemid（第34行）。