在 php 中获取请求 url 的 Content-Type

Question

Using php how would I be able to define the variable $type into the content-type of http://www.example.com

For example: $type defined as "text/html"

So far this is what i am working with:

<?php
$url = 'http://www.example.com';
print_r(get_headers($url));
print_r(get_headers($url, 1));
?>

The code may be changed as much as needed

Answer 1

Have you tried:

$type = get_headers($url, 1)["Content-Type"];

As noted in comments by @Michael, this syntax won't work without a very current version of PHP.

Have you also tried:

$headers = get_headers($url, 1);
$type = $headers["Content-Type"];

?

Answer 2

Sometimes get_header return wrong values becouse it read http headers, but not file. It should be better use finfo:

$finfo = new finfo(FILEINFO_MIME_TYPE);
$type = $finfo->buffer(file_get_contents($link));

Answer 3

Not very clear on what you are trying to do, but if you are trying to get the request content type that was sent by the browser to your script, you can do this:

<?php 
// Collect all headers
$headers = [];
foreach(getallheaders() as $name => $header){
    $headers[strtolower($name)] = $header;
}

// Get the content type header
$contentType = $headers['content-type'];
?>