Print specific array value via function argument

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.)

It's not clear what you're trying to accomplish. Could you clarify more on "I'm stuck at the point on how to pass the $key value trough the function."? — Amal
– Amal, Commented Mar 7, 2014 at 16:39

AbraCadaver · Accepted Answer · 2014-03-07 16:48:16Z

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.

edited Mar 7, 2014 at 16:48

answered Mar 7, 2014 at 16:43

AbraCadaver

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1 Comment

Rico Over a year ago

Aah, yes ofcourse! I had it a for loop first where i used $menu[$i], i guess i didn't think it trough when i changed it. Your first solution was what i was looking for, thanks!

BT643 · Accepted Answer · 2014-03-07 16:44:34Z

0

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); ?>

answered Mar 7, 2014 at 16:44

BT643

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Comments

nanobash · Accepted Answer · 2014-03-07 16:47:57Z

0

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]

edited Mar 7, 2014 at 16:47

answered Mar 7, 2014 at 16:41

nanobash

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3 Comments

Rico Over a year ago

If echo out $menu it shows all entries. I want to pick out specific entries to print.

nanobash Over a year ago

@Rico If you want to make specific links than you have to parse them in loop, update your question and I'll help you

nanobash Over a year ago

@Rico I meant, explain what do you mean in specific links ? show me desirable output

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