我如何检查一个表中是否存在会话ID，而另一表中是否存在该ID，则回显错误或成功代码

Question

如何检查一个表列中是否存在会话ID，而另一表列中是否存在该会话ID，并回显错误或成功代码。

我已经尝试了以下内容，但没有回显任何内容

<?php
$result = $DBcon->query("SELECT username, email, Phone_number, subcription  FROM tbl_users join mergeing on tbl_users.user_id = mergeing.donator_1 where mergeing.donor  = {$_SESSION['userSession']}");
if(!$result){
    echo "user does not exist in Db";
}
else{
    while($row = $result->fetch_array()){
    }
}
?>

Answer 1

我用这个解决了

$result = $DBcon->query("SELECT username, email, Phone_number, subcription FROM tbl_users join mergeing on tbl_users.user_id = mergeing.donator_1 where mergeing.donor  = {$_SESSION['userSession']}");
if ($result->num_rows == 0) {
    echo "YOU HAVE NOT YET BEEN CONFIRMED";
}else{
while($row = $result->fetch_array())
{
echo $row['username']; 
}