php function not returning an array - php

Question

I have the following code, sorry for the length but it saves explaining most of what i need...
the $result in the pages() function is not being returned. When I call for it all I get is undefined variable. Why?
What i need is to pass the $start and $display variables to the mysql query.
<?php
if(isset($_POST['get_records_submit'])) {
$pages; $start; $display; $result;
function pages($stmt) {
//set how many records per page
$display = 10;
//determine how many pages there are
if(isset($_GET['p']) && is_numeric($_GET['p'])) { //already determined
$pages = $_GET['p'];
}
else {
$records = mysqli_stmt_num_rows($stmt);
}
//calculate the number of pages
if($records > $display) { //if there are more records than will fit on one page
$pages = ceil($records/$display); //ceil() rounds up to the nearest integer
}
else { //records will fit on one page
$pages = 1;
}
//determine which row to start returning results from
if(isset($_GET['s']) && is_numeric($_GET['S'])) { //already determined
$start = $_GET['s'];
}
else {
$start = 0;
}
$result = array(0=>$display, 1=>$pages , 2=>$start);
return $result;
}
$searchby = $_POST['search_by'];
$searchfor = $_POST['search_for'];
$contact = 1;
$i = 0;
//set the initial query
$query = "SELECT client_id, title_desc, firstname, surname, house, street, town, city, county, postcode as count FROM address LEFT JOIN client USING (client_id) LEFT JOIN client_title USING (title_id) LEFT JOIN address_street USING (address_id) LEFT JOIN address_town USING (address_id) LEFT JOIN address_city USING (address_id) LEFT JOIN address_county USING (address_id) WHERE is_contact = ?";
//depending on search terms, amend the query
if($searchby == 'all') {
$query .= " ORDER BY surname ASC, firstname ASC";
$stmt = mysqli_prepare($dbc, $query);
mysqli_stmt_bind_param($stmt, 'i', $contact);
mysqli_stmt_execute($stmt);
mysqli_stmt_store_result($stmt);
pages($stmt);
var_dump ($result);
foreach ($result as $var) { echo $var.' ';}
mysqli_stmt_close($stmt);
$query .= " ORDER BY surname ASC, firstname ASC LIMIT ?, ?";
$stmt = mysqli_prepare($dbc, $query);
mysqli_stmt_bind_param($stmt, 'iii', $contact, $start, $display);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $client_id, $stitle, $sfname, $ssname, $shouse,$sstreet, $stown, $scity, $scounty, $spostcode);
if($searchfor != '') {
echo "<p>You searched under <span class=\"bold\">\"All\"</span>, therefore your search term <span class=\"bold\">\"$searchfor\"</span> has not been included.</p>";
}
}
}

Answer 1

In this line:
pages($stmt);
The function pages() returns the array, but this result is not being set to a variable. Try using:
$result = pages($stmt);
And then accessing the array via the variable $result.
That $result variable that you declared up top? The reason it's not being set is because the function pages() doesn't have access to it because it's not within the function's scope.
Edit
As Lyth pointed out, if you don't want the function to return anything, you can modify your function as follows:
function pages($stmt) {
global $result;
// ...
// The rest of your function
// ...
$result = array(0=>$display, 1=>$pages , 2=>$start);
// Remove the 'return $result' line
}

Answer 2

I do not know why you write $pages; $start; $display; $result;? If you want to get the value of array try to use this...
$result = array($display, $pages , $start);
return $result;
You can check the array with
print_r ($result);