Question

我有两张桌子。首先是产品，第二个是订单

产品

id |Product name

01 |Computer
02 |Mobile
03 |Tablet

和订单的第二个表

p_id | username

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

现在我想向用户展示那些尚未被用户购买的产品假设登录用户名是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;

以上查询计算用户名不是mike的所有订单并显示结果..我只想要那些“Mike”未购买的产品列表

Answer 1

您必须进行反加入

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

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

的更多信息

Answer 2

我不确定我是否理解你的问题，但我认为你想要这样做：

$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

我想这就是你的目标。

$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);

Answer 4

你应该选择* FROM product WHERE id NOT IN（SELECT p_id FROM order WHERE user_name =＆＃39; Mike＆＃39 ;）那就是

在联合查询中获取id

4 个答案: