从用户表中获取朋友的个人资料图片

Question

I have 2 table's:

Users (id, username, email, avatar, etc...);

Friends (id, user1, user2, status);

Now I want to build on my profile page an list of my friends with there avatar(s). I'm trying for like 4 hours by myself but i don't get it... :(

BTW: this is an error i got!

Notice: Array to string conversion in /home/reduaqi158/domains/reduankurtaj.eu/public_html/snapfriends/vrienden.php on line 26

This is what i have right now:

<?php

    error_reporting(E_ALL);
    session_start();
    $username = $_SESSION['username'];
    $status = 2;

    include "includes/conn.php";

    $vrienden=mysqli_query($server,"SELECT * FROM vrienden WHERE status='$status' && vriend1='$username' || vriend2='$username' ");

    $vriend_list = array();
        while($row = mysqli_fetch_array($vrienden))
        {
            if ($row['vriend1'] == $username) {
                $vriend_list[] = $row['vriend2'];
            }
            else {
                $vriend_list[] = $row['vriend1'];
            }
        }

        echo json_encode($vriend_list);


    $foto=mysqli_query($server,"SELECT prof_pic FROM users WHERE username='$vriend_list['vriend1''vriend2']' ");
        while($row2 = mysqli_fetch_array($foto)) {

            echo "<img class='img-rounded' src=assets/profiel/".$row2['prof_pic']." alt='Card image cap'>";
        }

?>

json_encode output:

["ja","amando"]

Someone who can help me pls :)

Answer 1

Your initial approach is very confusing.

Almost everything in your code can be substituted by single SQL query.

You can use JOIN to get all your friends with their avatars in one go:

SELECT u.username as username, u.avatar as avatar,.... <== all columns which you need
    FROM `friends_table` f  <== your friends table
    JOIN `users_table` u    <== your users table
        ON (f.user1 = u.id) <== notice that i join on user1 column
    WHERE u.username = '$username' && f.status = '$status'
UNION
SELECT u.username as username, u.avatar as avatar,.... <== same columns
    FROM `friends_table` f  <== your friends table
    JOIN `users_table` u    <== your users table
        ON (f.user2 = u.id) <== notice that i join on user2 column
    WHERE u.username = '$username' && f.status = '$status'

By this query you select all users who are in a friendship with your $username. You need union because you don't know in which field ( user1 or user2 ) your $username is located.

NOTE: I strongly suggest using prepared statements instead of just putting '$var' inside SQL query to prevent SQL Injection.

After executing this query you can parse results and display avatars in such a way:

while($row = mysqli_fetch_array($vrienden, MYSQLI_ASSOC))
{
    echo "<img class='img-rounded' src=assets/profiel/".$row['avatar']." alt='Card image cap'>";
}

I hope you got the idea.

Answer 2

in your while statement you have to declare a value for the array. like array[0] = value. so that you know that array position 0 has a certain value. Like what I did here below. Don't know if it's in PHP like this but certain in .net you have to declare the location of a value in an array.

while($row = mysqli_fetch_array($vrienden))
    {
        if ($row['vriend1'] == $username) {
            $vriend_list[0] = $row['vriend2'];
        }
        else {
            $vriend_list[1] = $row['vriend1'];
        }
    }

and the following

$foto=mysqli_query($server,"SELECT prof_pic FROM users WHERE username='$vriend_list['vriend1''vriend2']' ");

shouldn't it be $vriend_list['vriend1'] . $vriend_list['vriend2']'

you have to use a connect character (the . in PHP)