从友谊页面中排除自己 - MySQL

Question

这是我获取用户朋友的查询。但有一个问题。如果接收者是会话用户，它将自己作为朋友返回。

SELECT user.name_surname, user.id, friendship.receiver_id, friendship.sender_id
FROM user
JOIN friendship ON user.id = friendship.receiver_id
WHERE 
    (friendship.receiver_id =".$_SESSION["SES_USER_ID"]." OR friendship.sender_id =".$_SESSION["SES_USER_ID"].") 
     AND is_approved='1'"

下表是我的MySQL表设计。我以user_id =16身份登录了网站。然后user_id=19向我发送了一份请求。然后我批准了。这里的一切都很美好但是当我想看到我的朋友时，我也将自己视为我的朋友。

如何在朋友页面上排除自己？

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尝试

如果我更改此行

JOIN friendship ON user.id = friendship.receiver_id

要

JOIN friendship ON user.id = friendship.sender_id

这次向我发送友情请求的另一位用户将自己视为他/她的朋友。

Answer 1

将

"AND user.id != " . $_SESSION["SES_USER_ID"]

不行吗？所以你的整个查询将是：

SELECT user.name_surname, user.id, friendship.receiver_id, friendship.sender_id
FROM user
JOIN friendship ON user.id = friendship.receiver_id
WHERE (friendship.receiver_id =".$_SESSION["SES_USER_ID"]." OR friendship.sender_id =".$_SESSION["SES_USER_ID"] . ") 
AND is_approved='1'
AND user.id != " . $_SESSION["SES_USER_ID"]

编辑： 我认为您的查询应该是：

SELECT user.name_surname, user.id, friendship.receiver_id, friendship.sender_id
FROM user
JOIN friendship ON user.id = friendship.receiver_id
AND friendship.receiver_id != " . $_SESSION['SES_USER_ID'] . "
WHERE (friendship.receiver_id =".$_SESSION["SES_USER_ID"]." OR friendship.sender_id =".$_SESSION["SES_USER_ID"] . ") 
AND is_approved='1'

Answer 2

SELECT user.name_surname, user.id, friendship.receiver_id, friendship.sender_id
FROM user
JOIN friendship ON user.id = friendship.sender_id OR user.id = friendship.receiver_id
WHERE (friendship.receiver_id =".$_SESSION["SES_USER_ID"]." OR friendship.sender_id  =".$_SESSION["SES_USER_ID"].") 
AND is_approved='1'
AND user.id != ".$_SESSION["SES_USER_ID"]."