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Question

I have a search input to search for the required subject. I am able to display the appropriate result, I want the result to be displayed but user can only to go to the link if he logins, else display message box "You have to login to be able to go to another page". The problem here is i can't seem to display the error message at all. I want to display the result even if user is not or is logged in. Just that directing him to another page display error message and he must login or he cant be directed

<form class="search">
                <div class="form-group input">
                <div class="icon-addon">
                <input type="text" name="type" class="st-search" id="search" placeholder="Search"  autocomplete="off" >
                <label for="email" class="glyphicon glyphicon-search" rel="tooltip" title="email"></label>
                </div>
                <h4 id="results-text">Showing: <b id="search-string">Output</b></h4>
                <ul id="results"></ul>                                  
        </div>      
</form>

find.php

<?php
$dbhost = "hostname";
$dbname = "databasename";
$dbuser = "root";
$dbpass = "";

global $task_db;

$task_db = new mysqli();
$task_db->connect($dbhost, $dbuser, $dbpass, $dbname);
$task_db->set_charset("utf8");

if ($task_db->connect_errno) {
    printf("You can't connect: %s\n", $task_db->connect_error);
    exit();
}
    session_start();
    $_SESSION['login'] = true;
    // If user is not logged in
    if(!$_SESSION['login']){
        $message = 'You are not logged in';
        echo $message;
   die;
    }
//if user is logged in, user can be redirected to the link. Something is not right here
else{
$html = '';
$html .= '<li class="result">';
$html .= '<a target="_blank" href="url">';
$html .= '<h3>name</h3>';

$html .= '</a>';
$html .= '</li>';

$search = preg_replace("/[^A-Za-z0-9]/", " ", $_POST['query']);
$search = $task_db->real_escape_string($search);

if (strlen($search) >= 1 && $search !== ' ') {
    $query = 'SELECT * FROM tablename WHERE columnname LIKE "%'.$search_string.'%"';
    $result = $task_db->query($query) or trigger_error($task_db->error."[$query]");
    while($results = $result->fetch_array()) {
        $result_array[] = $results;
    }
    if (isset($result_array)) {
        foreach ($result_array as $result) {
            $name = preg_replace("/".$search_string."/i", "<b class='highlight'>".$search."</b>", $result['subject_Name']);
            //display url
            $url = 'https://www.google.com/'.urlencode($result['subject_Name']).'&lang=en';
            $output = str_replace('name', $name, $html);
            $output = str_replace('url', $url, $output);
            echo($output);
        }
    }else{
        // No Results found
    }
}
}
?>

Answer 1

First session_start(); shoul be the first command Second

 $_SESSION['login'] = true;

sets $_SESSION['login'] allway to true so that

if(!$_SESSION['login']){

will never happen.

Answer 2

Yes, because you set $_SESSION['login'] = true;
so the check if (!$_SESSION['login']){ will never be true.

You always get the result;

//$_SESSION['login'] = true;

When your user login set the $_SESSION['login'] = true; but not just before the if (!$_SESSION['login']){ line