在登录时显示当前用户信息

Question

I am trying to display the current user info when only they are logged in.. I have got the users info to display but it displays all users not just the logged in user. it is showing all users

<?php
        $sql = "SELECT users.uid, users.username, user_images.user_image
        FROM users, user_images
        WHERE users.user_id = users.user_id";

        $result = query($sql);
        while (($row = mysqli_fetch_assoc($result)) != false) {

            $uid = $row['uid'];
            $user_image = $row['user_image'];
            $username = $row['username'];


            echo "$username";
            echo "$user_image";

        }        
 ?> 

My Login

function login_user($email, $password)
{
    $sql = "SELECT user_pwd, uid FROM users WHERE user_email = '" . escape($email) . "' AND active = 1";

    $result = query($sql);

    if (row_count($result) == 1) {

        $row = fetch_array($result);

        $db_password = $row['user_pwd'];

        if (password_verify($password, $db_password)) {

            $_SESSION['email'] = $email;            

            return true;
        } else {

            return false;
        }

        return true;
    } else {

        return false;
    }
}

Query Functions

function row_count($result)
{
    return mysqli_num_rows($result);
}


function query($query)
{
    global $connection;
    return mysqli_query($connection, $query);
}


function fetch_array($result)
{
    global $connection;
    return mysqli_fetch_array($result);
}

Answer 1

In your login you should create a session variable for userid

 function login_user($email, $password)
    {
        $sql = "SELECT user_pwd, uid FROM users WHERE user_email = '" . escape($email) . "' AND active = 1";

        $result = query($sql);

        if (row_count($result) == 1) {

            $row = fetch_array($result);

            $db_password = $row['user_pwd'];

            if (password_verify($password, $db_password)) {

                $_SESSION['email'] = $email;            
                $_SESSION['user_id'] = $row['user_id'];
                return true;
            } else {

                return false;
            }

            return true;
        } else {

            return false;
        }
    }

then in the php file where you want to display the users info you will use $_SESSION['user_id']

<?php
        session_start();
        $user_id = $_SESSION['user_id'];
        $sql = "SELECT users.uid, users.username, user_images.user_image
        FROM users, user_images
        WHERE users.user_id = {$user_id}";

        $result = query($sql);
        while (($row = mysqli_fetch_assoc($result)) != false) {

            $uid = $row['uid'];
            $user_image = $row['user_image'];
            $username = $row['username'];


            echo "$username";
            echo "$user_image";

        }        
 ?> 

Hope it helps.