如何仅显示之前与 [PHP] 聊天过的用户

Question

我有table个用户

**| 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（用户名1）的回复，它将显示

username 1
username 1
username 2

我需要展示一下

username 1
username 2

能不能请教一下，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']

如何仅显示之前与 [PHP] 聊天过的用户

how to show only users who have previously chat with [PHP]

php

mysql

sql

chat