PHP/MYSQL echo logged in user's data from database table

Question

can anyone help me with the situation I'm in. I've been trying to find answers on how to display a user's first name, last name email...etc from the table that has the info stored but my only luck with echoing this info is only getting either ALL the info for a certain feel to show like all emails instead of just 1 specific email for the user or I just end up with mysql errors.

here's my index.php source

<?php

session_start();

if (isset($_SESSION['id'])) {
  // Put stored session variables into local php variable
  $userid = $_SESSION['id'];
  $email = $_SESSION['email'];
  $firstname = $_SESSION['first_name'];
  $lastname = $_SESSION['last_name'];
  $businessname = $_SESSION['company_name'];
  $country = $_SESSION['country'];
  $plan = $_SESSION['plan'];
} else {
    header("Location: http://somewebsite.com");
}   

include 'connect.php';

$first_name = $_GET['first_name'];
$last_name = $_GET['last_name'];

?>

for instance I would like to echo the user's first name in the header

<span class="username">USER</span> <--replace user with logged in user's firstname n last        name-->

Where are you accessing the database? You said you tried it and received errors. Please show the code! — Tobias Golbs
– Tobias Golbs, Commented Aug 30, 2013 at 7:39
I don't understand what is the problem here. Please explain more and show the errors you get. — iiro
– iiro, Commented Aug 30, 2013 at 8:43
If you’re saving your user data to a session, why can’t you then echo the user’s first name? What’s the problem? — Martin Bean
– Martin Bean, Commented Aug 30, 2013 at 9:35
You need to post your code that accesses database. Otherwise we cannot say, what is wrong with your code. — Lukas1
– Lukas1, Commented Aug 30, 2013 at 10:27

Vikas Arora · Accepted Answer · 2013-08-30 10:21:46Z

1

You can do something like this: Write a function to fetch specific data from the table such as one below.

function getuserfield($field) {
    $query = "SELECT $field FROM users WHERE id='".$userid."'";
    if ($query_run = mysqli_query($query)) {
        if ($query_result = mysqli_result($query_run, 0, $field)) {
            return $query_result;
        }
    }
}
/* userid = $_SESSION['id']; */

The "$query_result" will display the username. $field is the specific field you want to display. By this function you can fetch any particular data from the specified field

edited Aug 30, 2013 at 10:21

answered Aug 30, 2013 at 7:52

Vikas Arora

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2 Comments

Tobias Golbs Over a year ago

Please do not recommend mysql_* - functions to the user. They are deprecated. Also your function uses $userid but does not declare it, neither is it passed as a parameter. Your code is also vulnerable for mysql injections. Really bad example!

Vikas Arora Over a year ago

@TobiasKun The userid variable is declared in the question above which is equal to $_SESSION['id']. I am using a function similar to this to display the current logged in user and other stats in my own website. I used mysql_ function by mistake..it should be mysqli_ function.

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