在IF語句中更新多個變量

Question

我在更新IF語句中的多個變量時遇到了一些麻煩，並且不確定哪裏出錯。

這裏是我的代碼：

$id = $_POST['retrive_quantity']; 
$n_quantity = $_POST['n_quantity']; 


$result2 = mysql_query ("SELECT * FROM stock_control WHERE id = '$id' "); 

while ($row1 = mysql_fetch_array($result2)) 
{ 
    $quantity=$row1['quantity']; 
    $threshold=$row1['threshold']; 
    $emailSent=$row1['email_sent']; 
    $orders_status=$row1['orders_status']; 
    $orders=$row1['orders']; 
} 

if ($quantity + $n_quantity > $threshold && $emailSent == 2) { 
    $update=0 && $orders_status=0 && $orders=0; 
} else { 
    $update=2 && $orders_status=1 && $orders=$orders; 
} 

mysql_query("UPDATE stock_control SET quantity=quantity + '$n_quantity', 
email_sent = '$update', orders_status = '$orders_status', orders = '$orders' 
WHERE id = '$id' "); 

if語句工作的第一線，並且還與變量的只有一個，但沒有多。

Answer 1

無法更新這樣你有;

if ($quantity + $n_quantity > $threshold && $emailSent == 2) { 
    $update=0; 
    $orders_status=0; 
    $orders=0; // or $update = $orders_status = $orders=0; 
} else { 
    $update=2; 
    $orders_status=1; 
    //$orders=$orders; //no need for this 
} 

Answer 2

給獨立的變量聲明，如果塊

$id = $_POST['retrive_quantity']; 
$n_quantity = $_POST['n_quantity']; 
$update=2; 
$orders_status=1; 
$orders=0; 
$result2 = mysql_query ("SELECT * FROM stock_control WHERE id = '$id' "); 

while ($row1 = mysql_fetch_array($result2)) 
{ 
    $quantity=$row1['quantity']; 
    $threshold=$row1['threshold']; 
    $emailSent=$row1['email_sent']; 
    $orders_status=$row1['orders_status']; 
    $orders=$row1['orders']; 
} 

if ($quantity + $n_quantity > $threshold && $emailSent == 2) { 
    $update=0; 
    $orders_status=0; 
    $orders=0; 
} else { 
    $update=2; 
    $orders_status=1; 
    $orders=$orders; 
} 

mysql_query("UPDATE stock_control SET quantity=quantity + '$n_quantity', 
email_sent = '$update', orders_status = '$orders_status', orders = '$orders' 
WHERE id = '$id' ");