Question

我有以下三种可能的网址..

www.mydomain.com/445/loggedin/?status=empty
www.mydomain.com/445/loggedin/?status=complete
www.mydomain.com/445/loggedin /

www.mydomain.com/445部分是动态生成的，每次都不同，所以我不能完全匹配，我怎样才能检测到以下内容......

如果$ url包含已登录但不包含/？status = empty OR /？status = complete

我尝试的一切都失败了，无论它总是会检测到登录的部分......

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

将网址分割成细分

$path = explode('/',$referrer);
$path = array_slice($path,1);

然后只需在该数组上使用您的逻辑，您包含的第一个URL就会返回：

Array ( [0] => 445 [1] => loggedin [2] => ?status=empty )

Answer 2

你可以这样做：

$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';
}

我建议查看array_intersect，这非常有用。

希望它有所帮助，不确定这是否是最佳方式，但可能会激发你的想象力。

Answer 3

Strpos可能不是您想要使用的。你可以用stristr：

来做

    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';
        }
    }

但使用正则表达式可能更容易/更好：

if(preg_match('/\/loggedin\/(\?status=(.+))?$/', $referrer, $match)) 
{
    if(count($match)==2) echo "The status is ".$match[2];
    else echo "The status is logged in";
}

if /和带有strpos的URL语句

3 个答案: