Can someone please tell me why this does not work?

Question

In an attempt to learn basic php, I wrote this program, but it only works if current user is set to "Mickey". Can someone please explain to me my error? Thank you.
<?php $current_user = "Mickey";
$firstnamearr = array (
    'Mickey', 
    'Minnie',
    'Donald',
     'Pluto',
     'Friend'
     );
$lastnamearr = array(
    'Mouse',
    'Mouse',
    'Duck',
    FALSE
    );
if($currentuser =='Mickey') {
    $fullname = $firstnamearr[0] . " " .  $lastnamearr[0];
    echo $fullname;
}elseif ($currentuser == 'Minnie') {
    $fullname = $firstnamearr[1] . " " . $lastname_arr[1];
}elseif ($currentuser  == 'Donald') {
    $fullname = $firstnamearr[2] . " " .  $lastnamearr[2];
}elseif ($currentuser  == 'Pluto') {
    $fullname = $firstnamearr[3] . " " .  $lastnamearr[3];
}else {
    $fullname = $firstnamearr[4] . " " .  $lastname_arr[4];
}
?>

Ida Brogie · Accepted Answer

maybe it is only when the first name is mickey you write it out with echo $full_name; 
you can put this line echo $fullname; last so it always will be written out. and you can put $fullname = ""; at the beginning if it won't go into any if-statements.

Kevin Korte · Answer

Yes, your problem is you only have an echo statement in the Micky block. When any other name gets put in, the full name variable gets generated, but it isn't put to the screen.
Also, just FYI, your final else statement will return an error. You're assigning $last_name to be equal to $last_name_arr[4] which is not in the array, so it errors.

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Laurel Natale

Laurel Natale

Can someone please tell me why this does not work?

2 Answers

Ida Brogie

Ida Brogie

Laurel Natale

Laurel Natale

Kevin Korte

Kevin Korte