using variables created in function

Question

I'm using simple html DOM and extracting two links from website.

function get_links($website) {
$html = file_get_html($website);   
$img = $html->find('div.entry img',0)->src;  
$url = $html->find('div.entry a',0)->href;}

how do I use $img and $url after I run function get_links?

lonesomeday · Accepted Answer · 2012-01-29 15:15:17Z

You have two options. The first is to return them:

function get_links($website) {
    $html = file_get_html($website);   
    $img = $html->find('div.entry img',0)->src;  
    $url = $html->find('div.entry a',0)->href;
    return array('img' => $img, 'url' => $url);
}

You can only return one value from a function, so you have to make an array.

The other option is to take arguments by reference:

function get_links($website, &$img, &$url) {
    $html = file_get_html($website);   
    $img = $html->find('div.entry img',0)->src;  
    $url = $html->find('div.entry a',0)->href;
}

When you call this, you can then provide two values, which will contain the values:

get_links($someurl, $image, $url);
echo $image; // echoes the image source
echo $url; // echoes the url

I expect the first technique (the array) would be the simpler.

You have other options: you could make $img and $url global. This is a Bad Idea. You could also define get_links as an anonymous function and use the use keyword, but this is less useful, I think. You could also encapsulate the function in an object:

class Links {
    public $url;
    public $img;

    function __construct($website) 
        $html = file_get_html($website);   
        $this->img = $html->find('div.entry img',0)->src;  
        $this->url = $html->find('div.entry a',0)->href;
    }
}

// elsewhere
$links = new Links($someurl);
$links->img;
$links->url;
}

xdazz · Accepted Answer · 2012-01-29 15:12:07Z

1

You can return them as an array:

return array(
    'img'  => $img,
    'url'    => $url
);

answered Jan 29, 2012 at 15:12

xdazz

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Comments

DonCallisto · Accepted Answer · 2012-01-29 15:13:50Z

0

There are different ways:

return them to caller (as an array)
declare them outside the function in a different scope optic
declare them globals
so on

Consider the solution that fit better to your goal.

answered Jan 29, 2012 at 15:13

DonCallisto

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Comments

Jeremy Harris · Accepted Answer · 2012-01-29 15:14:13Z

0

Your function is only partially complete. You need to return those values to the calling origin for use there. When returning multiple values from a function, I prefer to return it as an associative array.

function get_links($website) {
   $html = file_get_html($website);   
   $img = $html->find('div.entry img',0)->src;  
   $url = $html->find('div.entry a',0)->href;}
   $results = array('img' => $img, 'url' => $url);
   return $results;
}

answered Jan 29, 2012 at 15:14

Jeremy Harris

24.7k13 gold badges84 silver badges133 bronze badges

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