if / and statement for URL with strpos - php

Question

I have the following three possible urls..
www.mydomain.com/445/loggedin/?status=empty
www.mydomain.com/445/loggedin/?status=complete
www.mydomain.com/445/loggedin/
The www.mydomain.com/445 part is dynamically generated and is different each time so I can't do an exact match, how can i detect the following...
If $url contains loggedin but DOES NOT CONTAIN either /?status=empty OR /?status=complete
Everything I try fails as no matter what it will always detect the logged in part..
if(strpos($referrer, '?status=empty')) {
echo 'The status is empty';
}
elseif(strpos($referrer, '?status=complete')) {
echo 'The status is complete';
}
elseif(strpos($referrer, '/loggedin/')) {
echo 'The status is loggedin';
}

Answer 1

Slice up the URL into segments
$path = explode('/',$referrer);
$path = array_slice($path,1);
Then just use your logic on that array, the first URL you included would return this:
Array ( [0] => 445 [1] => loggedin [2] => ?status=empty )

Answer 2

You could do something like this:
$referrer = 'www.mydomain.com/445/loggedin/?status=empty';
// turn the referrer into an array, delimited by the /
$url = explode('/', $referrer);
// the statuses we check against as an array
$statuses = array('?status=complete', '?status=empty');
// If "loggedin" is found in the url, and count the array_intersect matches, if the matches = 0, none of the statuses you specified where found
if( in_array('loggedin', $url) && count(array_intersect($url, $statuses)) == 0 )
{
echo 'The user is logged in';
}
// if the complete status exists in the url
else if( in_array('?status=complete', $url) )
{
echo 'The status is complete';
}
// if the empty status exists in the url
else if( in_array('?status=empty', $url) )
{
echo 'The status is empty';
}
I would recommend looking at array_intersect, it is quite useful.
Hope it helps, not sure if this is the best way of doing it, but might spark your imagination.

Answer 3

Strpos is probably not what you want to use for this. You could do it with stristr:
if($test_str = stristr($referrer, '/loggedin/'))
{
if(stristr($test_str, '?status=empty'))
{
echo 'empty';
}
elseif (stristr($test_str, '?status=complete'))
{
echo 'complete';
} else {
echo 'logged in';
}
}
But it's probably easier/better to do it with regular expressions:
if(preg_match('/\/loggedin\/(\?status=(.+))?$/', $referrer, $match))
{
if(count($match)==2) echo "The status is ".$match[2];
else echo "The status is logged in";
}