I created a chrome extension and from popup.js I called PHP script (Using Xhttprequest) that reads the cookie. Like this:
$cookie_name = "mycookie";
if(isset($_COOKIE[$cookie_name]))
{
echo $_COOKIE[$cookie_name];
}
else{
echo "nocookie";
}
But I'm getting this warning at errors in extensions.
A cookie associated with a cross-site resource at (Here is my domain) was set without the
SameSite
attribute. A future release of Chrome will only deliver cookies with cross-site requests if they are set withSameSite=None
andSecure
. You can review cookies in developer tools under Application>Storage>Cookies and see more details at https://www.chromestatus.com/feature/5088147346030592 and https://www.chromestatus.com/feature/5633521622188032.
I tried to create a cookie like this but it didn't help.
setcookie($cookie_name,$cookie_value, time() + 3600*24, "/;samesite=None ","mydomain.com", 1);
Following instructions from this question.
I'm also in a "trial and error" for that, but this answer from Google Chrome Labs' Github helped me a little. I defined it into my main file and it worked - well, for only one third-party domain. Still making tests, but I'm eager to update this answer with a better solution :)
EDIT: I'm using PHP 7.4 now, and this syntax is working good (Sept 2020):
If you have PHP 7.2 or lower (as Robert's answered below):
setcookie('key', 'value', time()+(7*24*3600), "/; SameSite=None; Secure");
If your host is already updated to PHP 7.3, you can use (thanks to Mahn's comment):
Another thing you can try to check the cookies, is to enable the flag below, which—in their own words—"will add console warning messages for every single cookie potentially affected by this change":
chrome://flags/#cookie-deprecation-messages
See the whole code at: https://github.com/GoogleChromeLabs/samesite-examples/blob/master/php.md, they have the code for
same-site-cookies
too.