Question

mysql场景：

表：posts => postId，userId postContent；表格：friends => friendId，userA，userB；

查询：

SELECT *
FROM posts
WHERE userId IN (SELECT * FROM friends WHERE userA = {$_SESSION['userId']} OR userB = {$_SESSION['userId']});

mysqli_fetch_assoc()期望参数1为mysqli_result，给定布尔值

查询中我缺少什么？

Answer 1

请尝试这个

SELECT *
FROM statements 
WHERE userId IN (SELECT userId FROM friends WHERE userA = {$_SESSION['userId']} OR userB = {$_SESSION['userId']});

Answer 2

编辑：有一个子查询

SELECT *
FROM posts 
WHERE userId IN (
    SELECT 

    CASE 
        WHEN userA = {$_SESSION['userId']} THEN userB
        ELSE userA
    END as user

    FROM friends WHERE userA = {$_SESSION['userId']} OR userB = {$_SESSION['userId']}
);

旧答案：

我搜索了一个更好的解决方案，但我认为它可以用于两个子查询：

SELECT *
FROM posts 
WHERE userId IN (SELECT userB FROM friends WHERE userA = {$_SESSION['userId']}) OR userId IN (SELECT userA FROM friends WHERE userB = {$_SESSION['userId']});

但是请考虑使用PDO来提高安全性

使用同一查询中的另一个选择从表中选择

2 个答案: