dropdown menu with pagination - php

Question

I have a dropdown filter which displays results directly from my database. I also have pagination which controls my pages. I am trying to get my dropdown to store results which at the moment i have had no success. my filter has 3 things: all, snake, reptile. When I click the option it displays the name e.g. snake. But when I press page 2 or 3 the result disappears and the dropdown goes back to all rather then stay on the selected option.
Okay update: when i click page 1,2 or 3 then the filter it shows correct results is there a way i can skip having to keep clicking the filter to show results on next page
So far I have tried: I have tried to create a session and also kept session start at the beginning which didn't help. I have also changed post to get when requesting the results. I have results set to two per page but it only displays two results when I click the option. As soon as I press 1,2,3, it displays no results. I have also tried using JavaScript to create a function getselected items and added a one click get selected item but still no results. What i want to happen: I click snake (which has 4 names in the database) I want it to display two names on page 1, and 2. Here is my base code:
$s = "SELECT DISTINCT name from users";
$result = $con->query($s);
echo "<form id='id' method='get' name='myform'>";
echo "<select name='fetch' class='filter' action='pagprofiletest.php'>";
echo "<option value='All'>All</option>";
if($result = mysqli_query($con, $s)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
$output = "<option value='".$row['name']."'"; //keeps select option on dropdown the same upon page refresh
if($_GET['fetch'] == $row['name']){
$output .= " selected='selected'";
}
$output .= ">".$row['name']."</option>";
echo $output;
}
echo "</select>";
echo "<input id='submit' name='submit' type='submit' value='submit' onclick='reload()'></input>";
echo "</form>";
}
}
?>
<?php
// $sql = "SELECT count(id) FROM users WHERE approved='1'";
$sql = "SELECT count(*) FROM users";
$query = mysqli_query($con, $sql);
$row = mysqli_fetch_row($query);
//here we have the total row count
$rows = $row[0];
//this is the number of results we want displayed per page
$page_rows = 2;
//this tells us the page number of last page
$last = ceil($rows/$page_rows);
//this makes sure last cannot be less then 1
if($last < 1){
$last =11;
}
//establish $pagenum variable
$pagenum = 1;
//get pagenum from url vars if it is present, else it is =1
if(isset($_GET['pn'])){
$pagenum = preg_replace('#[^0-9]#', '', $_GET['pn']);
}
// this makes sure page number isnt below 1, or more then $last page
if($pagenum < 1){
$pagenum = 1;
} else if($pagenum > $last){
$pagenum = $last;
}
$limit = 'LIMIT ' .($pagenum - 1) * $page_rows .',' .$page_rows;
$sql = "SELECT name FROM users WHERE approved='1' $limit";
$query = mysqli_query($con, $sql);
$textline1 = "users (<b>$rows</b>)";
$textline2 = "Page <b>$pagenum</b> of <br>$last</b>";
$paginationCtrls = '';
if($last != 1){
if($pagenum > 1){
$previous = $pagenum -1;
$paginationCtrls .= 'previous ';
for($i = $pagenum-4; $i < $pagenum; $i++){
if($i > 0){
$paginationCtrls .=''.$i.' ';
}
}
}
$paginationCtrls .=''.$pagenum.' ';
for($i = $pagenum+1; $i <=$last; $i++){
$paginationCtrls .= ''.$i.' ';
if($i >= $pagenum+4){
break;
}
}
if($pagenum != $last){
$next = $pagenum + 1;
// $paginationCtrls .= ' Next ';
$paginationCtrls .= ' Next ';
}
}
if($pagenum != $last){
$lastpage= $last;
$paginationCtrls .= ' Last ';
}
if($pagenum == $last){
$first =l;
$paginationCtrls .= ' First ';
}