php – MySQL – NOT IN产生不需要的结果

我有以下表格：

user
+-----------------------------------------------+
|  user_id    |  username   | Password | ...    |
+-----------------------------------------------+
|     1       |     a       |  ***     | ...    |
+-----------------------------------------------+
|     2       |     b       |  ***     | ...    |
+-----------------------------------------------+
|     3       |     c       |  ***     | ...    |
+-----------------------------------------------+
|     4       |     d       |  ***     | ...    |
+-----------------------------------------------+
|     5       |     e       |  ***     | ...    |
+-----------------------------------------------+

friends
+-----------------------------------------------+
|  f_id    |  user_id   | friend_id | ...       |
+-----------------------------------------------+
|     1    |     4      |  2        | ...       |
+-----------------------------------------------+
|     2    |     4      |  1        | ...       |
+-----------------------------------------------+
|     3    |     4      |  5        | ...       |
+-----------------------------------------------+
|     4    |     4      |  3        | ...       |
+-----------------------------------------------+

我想让所有可用的用户都被添加为朋友(在这种情况下,user_id为1将有3个朋友要添加(2,3,5).但是,通过使用下面的SQL语句我只获得1个用户(4)可供添加：

$sql = "SELECT * FROM user WHERE user.user_id NOT IN 
(SELECT friends.friend_id FROM friends) AND 
user.user_id <> $_SESSION['id']." ORDER BY RAND() LIMIT 5";

但是当我以用户4登录时没有用户可以添加,这种方法很有效.这对我来说有点棘手.任何想法都将非常感激.

谢谢

解决方法:

编辑：

这个怎么样：
(in else子句使用无效的id或与登录用户相同的id)

select * from users where id not in
(
   select (
            case
              when uid = 1 then id
              when fid = 1 then uid
             else 0
            end
          ) from friends where uid = 1 or fid = 1
) and id != 1 order by rand() limit 5;

http://sqlfiddle.com/#!2/12de1/62

解决此问题的另一种方法(但可能不是最佳解决方案)：

如果用户拥有的朋友数量少于系统中的用户总数,则与两个表之间的完全连接相比,下面的联合查询可能不是昂贵的查询.

另外不要忘记在朋友表的uid和fid列上添加索引.

select * from users where id not in 
(
     select id from friends where uid = 1
     union
     select uid from friends where fid = 1
) and id != 1 order by rand() limit 5;

http://sqlfiddle.com/#!2/12de1/40

标签：notin,not-exists,php,mysql,join
来源： https://codeday.me/bug/20190901/1782381.html