Ggtcf

I had my web application and it has (registration form, adding info, viewing info, and search and all these choices in the same page in navbar menu )
I start session for each user(when log in) and every thing is OK till I search for something and I got the search result and I go to the add or view info I found that is logged out that is mean the session destroyed why this is happened ? I want user to be logged in even after I finish search pleas help me and apologize if I have mistaken.
this is my code for the search part :

<form class="navbar-form navbar-left" action="" method="post">
 <div class="input-group">
 <input size=55 type="text" name="search" class="form-control" placeholder="Search">
 <span class="input-group-btn">
 <button type="submit" class="btn btn-default"><span class="glyphicon glyphicon-search"></span></button>
 </span>
 </div>
 </form>


 </nav>
</div>

 <?php
if(!isset($_SESSION))session_start();
 $conn = mysql_connect("localhost", "root", '');

 mysql_select_db("db103",$conn);
$user=$_SESSION["myname"];
 $fname = '';

if (isset($_POST['search']))

 $fname = $_POST['search'];

else
if(isset($_SESSION["myname"]))
 $sql = "SELECT * FROM telephone_guide where owner='$user' and (firstName like '%$fname%' or secondName like '%$fname%' or phonenumber like '%$fname%' or celnumber like '%$fname%')";


 $result = mysql_query($sql, $conn) ;

// number of rows fetched
$num = mysql_num_rows($result);

echo("<div class='styl'>");
echo("<table width=300 border=3 >");

echo("<th>ID<th>First Name<th>Second Name<th>tele-Phone Number<th>cel_phone number<th>Date<th>Address");


 while ($arr = mysql_fetch_array($result)) 


echo("<tr><td><a href='update1.php?id1=$arr[0]'>$arr[0]</td><td>$arr[2]</td><td>$arr[3]</td><td>$arr[4]</td><td>$arr[5]</td><td>$arr[6]</td><td>$arr[7]</td></tr>");


 
echo("</table>");

?>
<br>
<br>
<?php
echo("</div>"); 

 ?>

You will need to initialize session before anything
You can also see ob_start() and ob_flush_end() at the very end of your script. This is to suppress errors arising from packets already sent due to intercalation html within php

So try This

<?php
ob_start();
session_start();
?>




<form class="navbar-form navbar-left" action="" method="post">
 <div class="input-group">
 <input size=55 type="text" name="search" class="form-control" placeholder="Search">
 <span class="input-group-btn">
 <button type="submit" class="btn btn-default"><span class="glyphicon glyphicon-search"></span></button>
 </span>
 </div>
 </form>


 </nav>
</div>

 <?php

if(isset($_SESSION))



 $conn = mysql_connect("localhost", "root", '');

 mysql_select_db("db103",$conn);
$user=$_SESSION["myname"];
 $fname = '';

if (isset($_POST['search']))

 $fname = $_POST['search'];

else
if(isset($_SESSION["myname"]))
 $sql = "SELECT * FROM telephone_guide where owner='$user' and (firstName like '%$fname%' or secondName like '%$fname%' or phonenumber like '%$fname%' or celnumber like '%$fname%')";


 $result = mysql_query($sql, $conn) ;

// number of rows fetched
$num = mysql_num_rows($result);

echo("<div class='styl'>");
echo("<table width=300 border=3 >");

echo("<th>ID<th>First Name<th>Second Name<th>tele-Phone Number<th>cel_phone number<th>Date<th>Address");


 while ($arr = mysql_fetch_array($result)) 


echo("<tr><td><a href='update1.php?id1=$arr[0]'>$arr[0]</td><td>$arr[2]</td><td>$arr[3]</td><td>$arr[4]</td><td>$arr[5]</td><td>$arr[6]</td><td>$arr[7]</td></tr>");


 
echo("</table>");

?>
<br>
<br>
<?php
echo("</div>"); 

 ?>
<?php ob_end_flush(); ?>