这似乎总是为 $item_result->num_rows 返回 1; 即使数据库中有 0 行。但是，如果一个项目存在，它会正确更新该行。我确定我的语法有问题，但我很难理解这个 mysqli。
$item_query = "SELECT COUNT(*) FROM `rel` WHERE `cart_id` = '".$cartId."' && `id_item` = '".$item_id."'";
$item_result = $mysqli->query($item_query) or die($mysqli->error.__LINE__);
if($item_result->num_rows==1) {
$item_query2 = "SELECT * FROM `rel` WHERE `cart_id` = '".$cartId."' && `id_item` = '".$item_id."'";
$item_result2 = $mysqli->query($item_query2) or die($mysqli->error.__LINE__);
$getOldItems = $item_result2->fetch_array();
$oldQty = $getOldItems['amount'];
$oldNotes = $getOldItems['notes'];
$newQty = $oldQty + $item_quantity;
$newNotes = $oldNotes . $item_notes;
$update_qty = $mysqli->query("UPDATE rel SET amount = '$newQty', notes = '$newNotes' WHERE `cart_id` = '$cartId' && `id_item` = '$item_id'");
if(!$update_qty){
printf("Errormessage: %s\n", $mysqli->error);
}
header('Location: ./ordernew.php');
} else {
$insert_cart_item = $mysqli->query("INSERT INTO rel (`email`, `cart_id`, `id_item`, `amount`, `notes`) VALUES ('$email', '$cartId', '$item_id', '$item_quantity', '$item_notes')");
if(!$insert_cart_item) {
printf("Errormessage: %s\n", $mysqli->error);
}
header('Location: ./ordernew.php');
}