Print specific array value via function argument - php

Question

$menu = array(
0 =>'top',
1 =>'photography',
2 =>'about'
);
<?php
function main_menu ($menu) {
$return = '<div class="menu_entry">' . PHP_EOL .'';
foreach( $menu as $key => $value)
{
$return .= '<a class="menu" href="index.php#' . $menu[$key] . '">' . $menu[$key] . '</a>' . PHP_EOL .'';
}
$return .= '</div>';
return $return;
}
?>
<?php echo main_menu($menu[1]); ?>
What i basically want to do is to pass a specific array value when i'm echoing out the menu.
I'm building a single page website with anchors and i want to pass value's so i can echo out the "top"-link.
I'm stuck at the point on how to pass the $key value trough the function.
**edit: I'm trying to print specific links. I want a function that is able to print out an link but i want to specify the link to print via the function argument.
for example:
<?php echo main_menu($key = '0'); ?>
result:
prints url: top
<?php echo main_menu($key = '2'); ?>
result:
prints url: photography
**
(A lack of jargon makes it a bit harder to explain and even harder to google.
I got my books in front of me but this is taking a lot more time than it should.)

Answer 1

You either need to pass the entire array and loop, or pass a single array item and not loop:
Single Item:
function main_menu ($menu) {
$return = '<div class="menu_entry">' . PHP_EOL .'';
$return .= '<a class="menu" href="index.php#' . $menu . '">' . $menu . '</a>' . PHP_EOL .'';
$return .= '</div>';
return $return;
}
echo main_menu($menu[1]);
Entire Array:
function main_menu ($menu) {
$return = '<div class="menu_entry">' . PHP_EOL .'';
foreach($menu as $value) {
$return .= '<a class="menu" href="index.php#' . $value . '">' . $value . '</a>' . PHP_EOL .'';
}
$return .= '</div>';
return $return;
}
echo main_menu($menu);
You don't need $menu[$key] just use the $value.

Answer 2

Should you not just be using $value inside your loop? And passing the entire array rather than one item of the $menu array?
$menu = array(
0 =>'top',
1 =>'photography',
2 =>'about'
);
<?php
function main_menu ($menu) {
$return = '<div class="menu_entry">' . PHP_EOL .'';
foreach( $menu as $key => $value)
{
$return .= '<a class="menu" href="index.php#' . $value . '">' . $value . '</a>' . PHP_EOL .'';
}
$return .= '</div>';
return $return;
}
?>
<?php echo main_menu($menu); ?>

Answer 3

Try:
echo main_menu($menu); // You will get your links printed
Instead of
echo main_menu($menu[1]); // In this case error is occured like : **Invalid argument supplied for foreach**
NOTE: You can use $value instead of $menu[$key]