建議與用戶不是朋友的朋友

Question

我想向用戶推薦朋友。我需要驗證，如果朋友想建議是不是朋友已經與用戶
我有2個表稱爲用戶，友誼

我試試這個代碼，但我無法繼續，怎麼我沒有更多的想法

$sql=「SELECT user.fullname,user.userID FROM users WHERE user. userID NOT 
    IN(SELECT * FROM friendship WHERE userID = '$userID') 「; 

我使用PHP和MySQL

Answer 1

嘗試用friendship.userID

$sql = "SELECT user.fullname,user.userID FROM users WHERE user. userID NOT 
IN(SELECT friendship.userID FROM friendship WHERE userID = '$userID')"; 

您需要檢查與使用擺脫掉那些已經在用戶的朋友

Answer 2

這可以幫助你..

function addFriend() { 

global $userid, $friendid; 

$check = mysql_query("SELECT * FROM friends WHERE userid = $userid AND friendid =  $friendid"); 

if (mysql_num_rows($check) == 1) { 
    exit("Youre already friend with this user"); 
} else { 

$sql = "INSERT INTO friends (userid, friendid) VALUES ($userid, $friendid)"; 
mysql_query($sql); 
if (mysql_affected_rows()) { 
    echo "Success"; 
else 
    echo "Failure"; 
} 
}