PHP不会回显内部联接查询的结果

Question

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

Answer 1

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."'";

Answer 2

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>";
    }