Question

I have this function

---

function theme_status_time_link($status, $is_link = true) {
    $time = strtotime($status->created_at);
    if ($time > 0) {
        if (twitter_date('dmy') == twitter_date('dmy', $time) && !setting_fetch('timestamp')) {
            $out = format_interval(time() - $time, 1). ' ago';
        } else {
            $out = twitter_date('H:i', $time);
        }
    } else {
        $out = $status->created_at;
    }
    if ($is_link)
        $out = "<a href='status/{$status->id}' class='time'>$out</a>";
    return $out;
}

and this

function twitter_is_reply($status) {
    $html = " <b><a href='user/{$status->from->screen_name}'>{$status->from->screen_name}</a></b><br /> $actions $link<br />{$text} <small>$source</small>";
}

I need to pass the variable $out from the first function to the second function, precisely to the $html variable in the second function. However, everything I try either gives me errors and outputs nothing. Without using globals because it appears multiple times in my script. Thanks.

Answer 1

If you want to use functions send it as an argument:

function twitter_is_reply($status, $html) {
    $html .= " <b><a href='user/{$status->from->screen_name}'>{$status->from->screen_name}</a></b><br /> $actions $link<br />{$text} <small>$source</small>";

    return $html;
}

Usage:

$out = theme_status_time_link();

echo twitter_is_reply('helloworld', $out);