How do I get the HTML code of a web page in PHP?

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I want to retrieve the HTML code of a link (web page) in PHP. For example, if the link is

https://stackoverflow.com/questions/ask

then I want the HTML code of the page which is served. I want to retrieve this HTML code and store it in a PHP variable.

How can I do this?

11 Answers

If your PHP server allows url fopen wrappers then the simplest way is:

$html = file_get_contents('https://stackoverflow.com/questions/ask');

If you need more control then you should look at the cURL functions:

$c = curl_init('https://stackoverflow.com/questions/ask');
curl_setopt($c, CURLOPT_RETURNTRANSFER, true);
//curl_setopt(... other options you want...)

$html = curl_exec($c);

if (curl_error($c))
    die(curl_error($c));

// Get the status code
$status = curl_getinfo($c, CURLINFO_HTTP_CODE);

curl_close($c);

Also if you want to manipulate the retrieved page somehow, you might want to try some php DOM parser. I find PHP Simple HTML DOM Parser very easy to use.

Simple way: Use file_get_contents():

$page = file_get_contents('http://stackoverflow.com/questions/ask');

Please note that allow_url_fopen must be true in you php.ini to be able to use URL-aware fopen wrappers.

More advanced way: If you cannot change your PHP configuration, allow_url_fopen is false by default and if ext/curl is installed, use the cURL library to connect to the desired page.

you can use the DomDocument method to get an individual HTML tag level variable too

$homepage = file_get_contents('https://www.example.com/');
$doc = new DOMDocument;
$doc->loadHTML($homepage);
$titles = $doc->getElementsByTagName('h3');
echo $titles->item(0)->nodeValue;

$output = file("http://www.example.com"); didn't work until I enabled: allow_url_fopen, allow_url_include, and file_uploads in php.ini for PHP7

I tried this code and it's working for me .

$html = file_get_contents('www.google.com');
$myVar = htmlspecialchars($html, ENT_QUOTES);
echo($myVar);
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