Question

我有两张桌子：

users：

USER_ID
USER_NAME
...

friends：

user_id（发送好友请求的用户）
friend_id（收到朋友请求的用户）
确认（1 =朋友，0 =朋友请求）

我希望每个用户都能看到他们的朋友和朋友请求。我有显示朋友用户名的问题。我使它工作，但这是一个糟糕的解决方案，我想只使用一个查询（如果可能）解决这个问题。

我试过了：

    $sql = "SELECT friends.friend_id, friends.user_id, users.user_name FROM friends
            INNER JOIN users ON friends.user_id = users.user_id
            WHERE friends.user_id = " . $_SESSION['user_id'] . " AND confirmed = 1
            OR friends.friend_id = " . $_SESSION['user_id'] . " AND confirmed = 1;";

我的问题是它会在好友表中显示user_name与user_id相同的user_id用户。我的查询需要检查user_id是否与$_SESSION['user_id']相同，如果相同则必须返回user_name与user_id相同的用户friend_id friends表格中的string date_time = "2018-04-24 10:00:00"; string[] words = date_time.Split(' ');//Split string string date = words[0];//date = 1st object (before space) string time = words[1];//time= 2nd object (after space)。

Answer 1

以下列方式制定您的SQL查询，

$sql = "SELECT 
    " . $_SESSION['user_id'] . " as user_id, 
    IF(u2.user_id = " . $_SESSION['user_id'] . ", u1.user_id, f.friend_id) as friend_id, 
    IF(f.user_id = " . $_SESSION['user_id'] . ", u2.user_name, u1.user_name) as friend_username
FROM users as u1
INNER JOIN friends as f
ON u1.user_id = f.user_id
INNER JOIN users as u2
ON f.friend_id = u2.user_id
WHERE (f.user_id = " . $_SESSION['user_id'] . " OR f.friend_id = " . $_SESSION['user_id'] . ") AND f.confirmed = 1";

这是现场演示： http://sqlfiddle.com/#!9/e10252/7

旁注：了解prepared statement因为您的查询现在容易受到SQL注入攻击。另请参阅how you can prevent SQL injection in PHP。

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