PHP Use returned value of a function in another Function

Question

The Following code is checking for links within a page and then trying to process those links and check if they work or not, by returning a header code. I basically need to use the returned value $links of the first function in the second one. is it possible?

here is my code

function checkPage ($content){ 
   $textLen = strlen($content);  
   $links = array (); 
   if ( $textLen > 5){ 
      $startPos = 0; 
      $valid = true; 

      while ($valid){ 
         $spos  = strpos($content,'<a ',$startPos); 
         if ($spos < $startPos) $valid = false; 
         $spos     = strpos($content,'href',$spos); 
         $spos     = strpos($content,'"',$spos)+1; 
         $epos     = strpos($content,'"',$spos); 
         $startPos = $epos; 
         $link = substr($content,$spos,$epos-$spos); 
         if (strpos($link,'https://') !== false) $links[] = $link;
         if (strpos($link,'http://') !== false) $links[] = $link;
      } 
   } 
  return **$links**; 
};

print_r(checkPage($content));

foreach ($links as &$link ) {
$ch = curl_init();
   curl_setopt($ch, CURLOPT_URL, $link);
   curl_setopt($ch, CURLOPT_HEADER, 1);
   curl_setopt($ch , CURLOPT_RETURNTRANSFER, 1);
   $data = curl_exec($ch);
   $headers = curl_getinfo($ch);
   if(curl_error($ch))
{
    echo 'error:' . curl_error($ch);
}   
   curl_close($ch);
   echo $link." returns code ".$headers['http_code']."<br />";  
};

return **$links**;? Do you mean that you simply want to store the value from the checkPage() function? This is PHP101... $links = checkPage($content);. — M. Eriksson
– M. Eriksson, Commented Jul 17, 2017 at 13:51
Do you just need $links = checkPage($content); ? That looks to be missing from your code, but that's very basic PHP considering what you already have. — iainn
– iainn, Commented Jul 17, 2017 at 13:55
Yes, I forgot to copy the variable "content" that contain a piece of HTML markup. Return $links; should be displayed without the asterisks as I tried to bold the word on the page editor here. — gcerni
– gcerni, Commented Jul 17, 2017 at 13:56
I would highly recommend going through some basic PHP-tutorials — M. Eriksson
– M. Eriksson, Commented Jul 17, 2017 at 13:57

David · Accepted Answer · 2017-07-17 13:56:45Z

2

You're printing the returned value, but then ignoring it after that:

print_r(checkPage($content));

Store the returned value in a variable. It doesn't need to be the same variable name. And in this case it probably shouldn't be the same variable name to keep the concept clear. So something like this:

$returnedLinks = checkPage($content);
print_r($returnedLinks);

foreach ($returnedLinks as $link ) {
    // do something with $link
}

Returning a value from a function doesn't make the variable itself available outside the function. It just means that when you call the function then that function call evaluates to a result, just as if you defined that result in-line instead of calling a function. You just need to store that result in a variable to use it, just like any other value.

answered Jul 17, 2017 at 13:56

David

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1 Comment

gcerni Over a year ago

Thanks, I'm still new around coding.

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