Get logged in username with php from database

Question

in the first php code !$fgmembersite->CheckLogin() works perfectly. also next in the html code <?= $fgmembersite->UserFullName(); ?>! works perfectly. But in the second php it won't work correctly. Am I doing something wrong in that code? if($row['name']=="$fgmembersite->UserFullName()")

<?PHP
require_once("./include/membersite_config.php");

if(!$fgmembersite->CheckLogin())
{
    $fgmembersite->RedirectToURL("login.php");
    exit;
}

?>  

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-US" lang="en-US">
<head>
      <meta http-equiv='Content-Type' content='text/html; charset=utf-8'/>
      <title>Home page</title>
      <link rel="STYLESHEET" type="text/css" href="style/fg_membersite.css">
</head>

<body>

<div id='fg_membersite_content'>

<h2>Home Page</h2>

Welcome back <?= $fgmembersite->UserFullName(); ?>!

</div>

</body>

<?php

include("db.php");
$select=mysql_query("select * from commenttable");
while($row=mysql_fetch_array($select))
{
if($row['name']=="$fgmembersite->UserFullName()")
{

        echo "<div id='sty'>";
        echo "<img src='files/fav icon.png'"."' width='50px' height='50px' align='left' />";
        echo "<div id='nameid'>".$row['name']."</div>";
        echo "<div id='msgid'>".$row['message']."</div>";
        echo "</div><br />";

}else{
ob_start();
        echo "<div id='sty'>";
        echo "<img src='files/fav icon.png'"."' width='50px' height='50px' align='left' />";
        echo "<div id='nameid'>".$row['name']."</div>";
        echo "<div id='msgid'>".$row['message']."</div>";
        echo "</div><br />";
ob_end_clean();
}
}

?>

Please reformat your question properly. See the preview and make sure it displays nicely before posting. Read: meta.stackexchange.com/editing-help — light
– light, Commented Jun 21, 2015 at 14:48

Professor of programming · Accepted Answer · 2015-06-21 15:19:05Z

1

You have an error with your PHP, you have enclosed what should be PHP code within a string causing it to output as you have typed.

Change the following line:

if($row['name']=="$fgmembersite->UserFullName()")

To the following:

if($row['name']==$fgmembersite->UserFullName())

edited Jun 21, 2015 at 15:19

answered Jun 21, 2015 at 14:50

Professor of programming

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