PHP Display option depending on value - php

Question

I am editing a part of a script that displays gender of a person but trying to add additional options.
The form that saves the information saves it as a value.
In my lang file this is what i have got related to the part i am trying to change :
$LNG['powner'] = 'Pet Owner';
$LNG['dog'] = 'Dog';
$LNG['cat'] = 'Cat';
$LNG['rabbit'] = 'rabbit';
The part of the code im trying to display is
((!empty($profile['gender'])) ? '<div class="sidebar-list">'.$LNG['ttl_gender'].':
<strong>'.(($profile['gender'] == 1) ? $LNG['powner'] :($profile['gender'] == 2) ?
$LNG['cat'] : ($profile['gender'] == 3) ? $LNG['rabbit'] : $LNG['rabbit']).'</strong>
</div>' : '').'
But it will only display cat or rabbit ...
I am thinking that the code is set to be one or the other because this works :
((!empty($profile['gender'])) ? '<div class="sidebar-list">'.$LNG['ttl_gender'].'
: <strong>'.(($profile['gender'] == 1) ? $LNG['dog'] : $LNG['cat']).'</strong></div>'
: '').
after the code if follows
'.((!empty($profile['website'])) ? '<div class="sidebar-list">'
.$LNG['profile_website'].': <strong><a href="'.$profile['website'].
'" target="_blank" rel="nofllow">'.$LNG['profile_view_site'].'</a>
</strong></div>' : '').'

Answer 1

Your code could be re-written to say:
if(!empty($profile['gender']) {
echo '<div class="sidebar-list">'.$LNG['ttl_gender'];
} else {
echo "<strong>";
if($profile['gender'] == 1) {
echo $LNG['powner'];
} else {
if($profile['gender'] == 2) {
echo $LNG['cat'];
}
else {
if($profile['gender'] == 3){
echo $LNG['rabbit'];
} else {
echo $LNG['rabbit'];
}
}
}
}
As you see the code using shorthand If/Else statements is confusing to read, and can lead to mistakes. Re-write your functionality in more readable way, and you will prevent errors, and confusion with regards to what the actual conditions should be.
I'm assuming that the problem you're having is that the users you're testing this on are either gender 2 or 3, and never 1.