php - mysql查询选择计数

Question

我必须使用计数和组，我有点困惑。

我有 2 张桌子：

• users

  id_user | 名字 | 姓氏 | 流派 | logo_src

和

• user_stalking

  用户 ID | user_stalking_id

目前，我的查询如下所示：

SELECT   us.user_id, us.user_stalking_id, us.notification_viewed, u.id_user,
         u.firstname, u.lastname, u.logo_src, u.genre, us.id, us.date
FROM     users u, user_stalking us
WHERE    us.user_id = ".$_SESSION['user']['id_user']."
     AND us.notification_viewed = 0
     AND us.user_stalking_id = u.id_user
ORDER BY us.date DESC

它返回：

约翰 | 琼森 | 图片1.png | 男性

现在我想接收在user_stalking.user_idanduser_stalking.user_stalking_id字段中包含相同 ID 的所有行的计数。

Answer 1

首先，我希望在此声明之前的某个地方你正在检查/清理 $_SEESION['user']['id_user'] 值，否则你会让自己对注入攻击敞开大门。

要回答您的问题，您可以通过像这样更改查询来检索这两个值（可能不是最有效的查询）。

SELECT  us.user_id, us.user_stalking_id, us.notification_viewed, u.id_user, u.firstname, u.lastname, u.logo_src, u.genre, us.id, us.date, 
    (SELECT COUNT(s.user_id) FROM user_stalking s WHERE s.user_id = u.user_id) AS stalking_count, 
    (SELECT COUNT(s2.user_id) FROM user_stalking s2 WHERE s2.user_stalking_id = u.user_id) AS being_stalked_count 
FROM    users u, user_stalking us 
WHERE   us.user_id = ".$_SESSION['user']['id_user']." 
    AND us.notification_viewed = 0 
    AND us.user_stalking_id = u.id_user ORDER BY us.date DESC

Answer 2

select count(*) AS count from users u, users_stalking us where u.user_id = us.user_stalking_id ??

（抱歉打错了，这是手机发的）

Answer 3

SELECT   COUNT(us.user_id)
FROM     users AS u
JOIN     user_stalking AS us ON (us.user_stalking_id = u.id_user)
WHERE    us.user_id = ".(int)$_SESSION['user']['id_user']."
          AND us.notification_viewed = 0
GROUP BY us.user_id
ORDER BY us.date DESC

Answer 4

“现在我想在 user_stalking.user_id 和 user_stalking.user_stalking_id 字段中接收包含相同 ID 的所有行的计数。”

我猜你正在寻找的是一个 equi join/inner join 查询。

    select count(*) from users u, user_stalking us where us.user_id = ".$_SESSION['user']['id_user']."
 AND u.id_user=us.user_stalking_id