如何仅显示以前与 [PHP] 聊天的用户

Question

我有表用户

**| username | password |**
username    1234
username1   1234
username2   1234

和消息

**| from_user_name | to_user_name | message |**
username        username1      Hi      
username        username2      Hi
username1       username      reply

这是我的SQL代码

$query = "SELECT * FROM " .table_name. " WHERE to_user_name = '".$_SESSION['username']."' OR from_user_name = '".$_SESSION['username']."' GROUP BY from_user_name,to_user_name";

我用(用户名(登录，如果有来自(用户名1(的回复，它将显示

username 1
username 1
username 2

我需要显示喜欢

username 1
username 2

你能告诉我吗，谢谢大家

Answer 1

使用：

SELECT DISTINCT CASE WHEN to_user_name = $_SESSION['username']
THEN from_user_name
ELSE to_user_name
END AS responder
FROM message
WHERE $_SESSION['username'] IN (from_user_name, to_user_name);

或

SELECT to_user_name AS responder
FROM message
WHERE from_user_name = $_SESSION['username']
UNION 
SELECT from_user_name 
FROM message
WHERE to_user_name = $_SESSION['username']