Access variable in function from another function. php

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.

Why do not you use reference: theme_status_time_link($status, $is_link = true, &$buffer){ ... } function twitter_is_reply($status, &$buffer) { .. } Is it what you think? By this you will be able to pass a reference to a variable. Tell me if this is your point. The second thing is you can call the second function in the first function, immediately before return. — Rolice
– Rolice, Commented Jul 12, 2011 at 21:37
can't you extends twitter_is_reply with a second parameter function twitter_is_reply($status, $out) {...} ? If else - provide more info how twitter_is_reply should be called... — madflow
– madflow, Commented Jul 12, 2011 at 21:39

Glass Robot · Accepted Answer · 2011-07-12 21:39:13Z

2

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);

answered Jul 12, 2011 at 21:39

Glass Robot

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Itay Moav -Malimovka · Accepted Answer · 2011-07-12 21:37:12Z

0

Now you start to understand why OO is so good.
Create one class, make the $out a member of it and the two functions, make them methods of that class.

answered Jul 12, 2011 at 21:37

Itay Moav -Malimovka

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1 Comment

Mob Over a year ago

If I understood any of the above, I wouldn't be asking this question.

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