php不会回应内部联接查询的结果

I've looked around the site for similar problems with solutions and non seem to fix the problem I have. I have two tables; "friendlist" and "users". I'm trying to use the "FriendID" from "friendlist" to retrieve information from the "users" table. Everything works fine up until the while($row = mysqli_fetch_array($result)){} loop then nothing else prints.

My code is as follows:

$query = "SELECT friendlist.FriendID, users.Name, users.Surname, users.Picture 
                FROM friendlist            
                INNER JOIN users
                ON friendlist.FriendID = users.Id
                WHERE friendlist.UserId ='".$id."'";
$result = mysqli_query($con, $query);

if(!$result){
    echo "<br/><h4>You currently do not have any friends. Please click the Find Friends button to find a friend</h4>";
}else{
    echo "<center><br/>Here is a list of all your friends:<br/>";
    echo "<table>";
    while($row = mysqli_fetch_array($result)){
        echo "<tr>";
        echo "<td>Pro Pic: <img style='width:200px; height:200px' alt='No Profile Picture' src='uploads/" .$row['Picture']. "' /></td>";
        echo "<td>Name :" .$row['Name']. "</td>";
        echo "<td>Surname :" .$row['Surname']. "</td>";
        echo "<td><form method='post' action='viewFriend.php'>";
        echo     "<input type='hidden' name='friendId' value='".$row['FriendID']."'/>";
        echo     "<input type='submit' name='View' value='View Profile'/>";
        echo "</form></td>";
        echo "</tr>";
    }
    echo "</table></center>";
}

Nothing is displayed on the browser. Only the level 4 heading text: "Here is a list of all your friends" shows. But after that its empty space.
I've checked the sql query on mySql and it works perfectly fine. I have no idea what's wrong. Any help will be much appreciated. Thanks

The problem is on your SQL Query. You did not put friendlist.UserId on the select part. That's why the where clause do not see where to compare it.

 $query = "SELECT friendlist.FriendID, users.Name, users.Surname, users.Picture 
                    FROM friendlist            
                    INNER JOIN users
                    ON friendlist.FriendID = users.Id
                    WHERE friendlist.UserId ='".$id."'";

Your code is correct but you miss the declaration of $id variable only. Use like below.

//initialize  the $id as.

$id = $_REQUEST['id'];

    $query = "SELECT friendlist.FriendID, users.Name, users.Surname, users.Picture 
                    FROM friendlist            
                    INNER JOIN users
                    ON friendlist.FriendID = users.Id
                    WHERE friendlist.UserId ='".$id."'";
    $result = mysqli_query($con, $query);

    if(!$result){
        echo "<br/><h4>You currently do not have any friends. Please click the Find Friends button to find a friend</h4>";
    }else{
        echo "<center><br/>Here is a list of all your friends:<br/>";
        echo "<table>";
        while($row = mysqli_fetch_array($result)){
            echo "<tr>";
            echo "<td>Pro Pic: <img style='width:200px; height:200px' alt='No Profile Picture' src='uploads/" .$row['Picture']. "' /></td>";
            echo "<td>Name :" .$row['Name']. "</td>";
            echo "<td>Surname :" .$row['Surname']. "</td>";
            echo "<td><form method='post' action='viewFriend.php'>";
            echo     "<input type='hidden' name='friendId' value='".$row['FriendID']."'/>";
            echo     "<input type='submit' name='View' value='View Profile'/>";
            echo "</form></td>";
            echo "</tr>";
        }
        echo "</table></center>";
    }