Php parse content from url by class name [closed]

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I can not find the right parsing solution from another link. I want to parse content from part of site.

I need get string "Director".

<div>
    <div class="jobtitle">Director</div>
</div>

This is code display all content from link

$url = 'somelink.com';
$result = file_get_contents($url);

echo($result);

Show us your parse() function and tell us what does not work as expected - actual vs. expected outcome. And let us know in detail what you've tried to debug your code. Please read How to Ask — jibsteroos
– jibsteroos, Commented Jun 24, 2020 at 18:54
Do you want to retrieve content of elements that has a certain class? Is that what you are saying? — bestprogrammerintheworld
– bestprogrammerintheworld, Commented Jun 24, 2020 at 18:59
@bestprogrammerintheworld thanks for library, best solution for my task. — stepbystep
– stepbystep, Commented Jun 25, 2020 at 21:32

EEAH · Accepted Answer · 2020-06-24 19:20:42Z

1

Try :

$url = 'somelink.com';
$result = file_get_contents($url);

$dom = new DOMDocument;
$dom->loadHTML($result);

$classname="jobtitle";
$finder = new DomXPath($dom);
$spaner = $finder->query("//*[contains(@class, '$classname')]");

Or if you don't want to use XPath traversal, you can simply loop throught all the div nodes and get the class attribute that is equal to "jobtitle".

This is the code:

$url = 'somelink.com';
$result = file_get_contents($url);

$dom = new DOMDocument;
$dom->loadHTML($result);

$nodes = $dom->getElementsByTagName ("div");

$wanted_node ;

foreach( $nodes $n) {
    if ($n->getAttribute('class') == "jobtitle"){
        $wanted_node = $n;
    }
}

//If wanted_node is not null(the node with class=jobtitle is found
if (isset ($wanted_node)){
    echo $wanted_node;
}

edited Jun 24, 2020 at 19:20

answered Jun 24, 2020 at 19:09

EEAH

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