PHP DOM从页面获取所有href并删除重复项

Question

I am creating a script that gets all links from an website and checking every link if is broken. My problem is, I need to display all links founded but I need to verify only the unique links not the duplicate. For example if a website has 4 links to google.com then I want to check only one time not four times.

foreach ($dom->getElementsByTagName('a') as $node) {

$info = $node->getAttribute( 'href' );

///The function that checks for broken links working. 
$check_url_status = check_url($info);

if ($check_url_status == '404') {

$badresult = 'Not working';

}else{

$badresult = 'Working';

}

$showlist .= '<li>The '.$info.' is '.$badresult.'</li>';

}


echo '<ul>'.$showlist.'</ul>';

This code is works, but I need to make it check the http status only 1 time for duplicate links.

I do not have idea of how to do this and also if is posible to do something like that.

Answer 1

You can make an array which you can use to save all already checked links in it. You will always check before checking the status if the link is already in the array or not. If so, skip link. If not, check status and add link to the array. You can skip an element by using the keyword continue.

$links = array();
foreach ($dom->getElementsByTagName('a') as $node) {
    $info = $node->getAttribute('href');

    if(!isset($links[$info])) {
        ///The function that checks for broken links working.
        $check_url_status = check_url($info);
        $links[$info] = $check_url_status;
    } else {
        $check_url_status = $links[$info];
    }

    if ($check_url_status == '404') {
        $badresult = 'Not working';
    } else {
        $badresult = 'Working';
    }

    $showlist .= '<li>The '.$info.' is '.$badresult.'</li>';
}

echo '<ul>'.$showlist.'</ul>';