Question

我没有得到任何数据，而且我已经检查了我的爸爸哪个应该匹配

我有两张表，likes和user_follow 我试图通过id绑定两个表。

表格 - 列

likes - idlikes，iduser，information 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，我应该能够显示与我关注的任何人相关的信息。

我没有得到任何错误，它只是没有回应任何东西？

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

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

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'

并获取值

$id_1 = $row_following['information'];

带有JOIN的MySQL查询未正确执行

1 个答案: