Problems returning an array in PHP

Question

Just trying to return an array in a function. Not having the best of luck...

( flag_friend_get_friends() is a function that returns a list of objects containing arrays )

So without a function, this works:

    <?php

    $users_friends = flag_friend_get_friends($user->uid);

      foreach ($users_friends as $id => $value) {     
        $users_friends_ids[] = $id;    
      }

    $test = $users_friends_ids;

    print $test[0];

    ?>

If I try and wrap it an a function, it doesn't work (nothing is printed)...:

         <?php

        function myfunc () {        

        $users_friends = flag_friend_get_friends($user->uid);

          foreach ($users_friends as $id => $value) {     
            $users_friends_ids[] = $id;    
            }

        return $users_friends;  

        }


        $test = myfunc();   

        print $test[0];

        ?>

What's the 'deliberate' mistake? :(

Updated code:

<?php
function myfunc () {        

$users_friends = flag_friend_get_friends($user->uid);
  foreach ($users_friends as $id => $value) {     
    $users_friends_ids[] = $id;    
    }

return $users_friends;  
}

$test = myfunc($user);  
print $test[0];
?>

Levi Hackwith · Accepted Answer · 2011-04-18 16:36:31Z

3

You need to pass $userinto your function when you call it:

$test = myfunc($user)

UPDATE:

<?php
function myfunc ($user) {        

$users_friends = flag_friend_get_friends($user->uid);
  foreach ($users_friends as $id => $value) {     
    $users_friends_ids[] = $id;    
    }

return $users_friends;  
}

$test = myfunc($user);  
print $test[0];
?>

That's what your code needs to look like.

edited Apr 18, 2011 at 16:36

answered Apr 18, 2011 at 13:27

Levi Hackwith

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2 Comments

james6848 Over a year ago

I'm afraid this doesn't seem to work. This is turning into a bit of a nightmare...

David Winant Over a year ago

for the $user parameter, you also need to declare it to the function, as in "function myfunc ($user) {"

David Winant · Accepted Answer · 2011-04-18 13:30:58Z

2

And you either need to pass in a reference for $users_friends_ids or declare it as global within the function.

The problem is that when you move stuff into a function it doesn't have access to your local variables any more.

answered Apr 18, 2011 at 13:30

David Winant

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2 Comments

Levi Hackwith Over a year ago

$users_friends_ids[] should work even when not declared properly inside the the function. In this case, PHP is just creating $users_friends_ids as an array and adding elements to it.

David Winant Over a year ago

it won't "work" because it won't be available outside of the function. granted, with the code above he's not accessing it outside of the function, but my point was that when you take code and move it into a function a common problem is that you don't have access to the same variable scope any more.

Coder1 · Accepted Answer · 2011-04-18 18:03:22Z

2

In drupal, $user is a global. So, you just need to define it at the top of your function ** if this function is always going to pull in the current user's friends.

function myfunc () {        
  // pull in the global $user var
  global $user;

  $users_friends = flag_friend_get_friends($user->uid);

  foreach ($users_friends as $id => $value) {     
    $users_friends_ids[] = $id;    
  }

  return $users_friends;  
}

answered Apr 18, 2011 at 18:03

Coder1

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