使用JOIN的MySQL查询无法正确执行

Question

我没有任何数据，我已经检查了我父亲应该匹配的数据

我有两个表， likes和user_follow我试图通过一个ID来绑定两个表。

表格-栏

likes赞-idlikes，iduser，信息user_follow -iduser_follow，iduser_follower，iduser_following

$following = $dbh -> prepare("SELECT L.* FROM likes L JOIN user_follow F ON F.iduser_following = L.iduser WHERE F.iduser_follower = ?");
$following->execute(array($_SESSION['user_auth']));
while($row_following = $following->fetch(PDO::FETCH_ASSOC)){
     $id_1 = $row_following['L.information']; // get id of user that i'm following
     echo $id_1;
}

因此，如果我关注某人，则我应该能够显示与我关注的人相关的信息。

我没有任何错误，只是没有回声什么？

---

样本数据

user_follow

iduser_follow           iduser_follower            iduser_following
     1                         2                        3
     2                         2                        4

likles

  idlikes                   iduser                  information
     1                         3                        info1
     2                         3                        info2

因此，假设$ _SESSION ['user_auth'] = 2，我应该输出info1和info2，对吗？

Answer 1

SELECT  b.*
FROM    user_follow a
        INNER JOIN likes b
            ON a.iduser_following = b.iduser
WHERE   a.iduser_follow = 'myuserID' AND
        iduser_following = 'followingID'

并获取价值

$id_1 = $row_following['information'];