在联合查询中获取ID

Question

我有两个表。第一产品和第二订货

产品

id |Product name 

01 |Computer 
02 |Mobile 
03 |Tablet 

和第二台订单

p_id | username 

02 | Joe 
02 | Mike 
01 | Joe 
03 | Tomy 

现在我想告诉这些产品的用户未通过用户尚未购买 让假设登录用户名是Mike

$username = "Mike"; 
$get = mysqli_query($db, "SELECT e.id 
FROM `products` AS e INNER JOIN `order` AS u ON u.p_id = e.id WHERE e.username != '$username' "); 
echo mysqli_num_rows($get); 

Result 3; 

以上查询计数其中username是不是迈克和显示结果的所有订单..我想只有那些没有被“迈克”购买

Answer 1

你必须做的那些产品列表上反连接

SELECT * FROM product p 
WHERE NOT EXISTS (
    SELECT 1 FROM order o WHERE o.username = 'Mike' AND p.id = o.p_id 
) 

更多信息https://dev.mysql.com/doc/refman/5.7/en/exists-and-not-exists-subqueries.html

Answer 2

我不知道如果我正确理解你的问题，但我认为要做到这一点：

$notThisUser = "Mike"; 
$sql = $mysqli->prepare("SELECT productname FROM product 
    WHERE id IN (SELECT p_id FROM order WHERE username != ?)"); 
$sql->bind_param("s", $notThisUser); 
$sql->execute(); 
$sql->bind_result($productName); 

while ($sql->fetch()) { 
    echo $productName; 
} 

Answer 3

你应该选择去* FROM product WHERE id NOT IN（SELECT p_id FROM order WHERE user_name = '迈克'）就是所有

Answer 4

我想这是你的目标是什么。

$username = "Mike"; 
$selectQuery = "select p.id from product 
inner join oder AS o on o.p_id=p.id where o.username <> :username"; 
$statement = $dbh->prepare($selectQuery); 
$statement->bindValue(":username", $username); 
$statement->execute(); 
$data = $statement->fetchAll(\PDO::FETCH_ASSOC);