Question

我有以下mysql查询：

$get_friends_posts = "SELECT * FROM users_friendships
                      WHERE userid1 != $myid and userid2 = $myid 
                      or userid2 != $myid and userid1 = $myid";

$friendpost = mysqli_query($mysql, $get_friends_posts);

$ myid是登录并观看此脚本的用户的唯一ID。现在，我希望通过一张支票或一些与我不相符的东西获得身份证，就像友谊的对手一样。

在数据库中有id，userid1和userid2的行。

Answer 1

此查询将从userid表中返回单个列（users_friendships），其中$myid为userid1或userid2。我假设没有userid1和userid2相同值的行。

SELECT
CASE WHEN userid1 = $myid THEN userid2 ELSE userid1 END AS userid
FROM users_friendships
WHERE userid1 = $myid OR userid2 = $myid

如果要使用这些结果从另一个表中获取数据，可以执行以下操作：

SELECT
posts.*
FROM posts
WHERE posts.userid IN
(
    SELECT
    CASE WHEN userid1 = $myid THEN userid2 ELSE userid1 END AS userid
    FROM users_friendships
    WHERE userid1 = $myid OR userid2 = $myid
)

或者，如果您需要有关用户和帖子的额外信息，则需要更复杂的需求，您可以执行以下操作：

SELECT
    users.name,
    posts.date,
    posts.title
FROM users_friendships
INNER JOIN users ON users.userid = CASE WHEN users_friendships.userid1 = $myid THEN users_friendships.userid2 ELSE users_friendships.userid1 END
INNER JOIN posts ON posts.userid = CASE WHEN users_friendships.userid1 = $myid THEN users_friendships.userid2 ELSE users_friendships.userid1 END
WHERE users_friendships.userid1 = $myid OR users_friendships.userid2 = $myid
ORDER BY posts.date DESC

如何获取与我的不匹配的mysql查询的id？

1 个答案: