用户如何向我的应用授予users_posts权限？

Question

I want to create a PHP application using the Facebook PHP SDK that pulls a users posts from the users wall and displays it as a feed on the users home page. According to the Graph API https://graph.facebook.com/{user-id}/feed is what I should request:

However, the API documentation says:

Your app needs user_posts permission from the person who created the post or the person tagged in the post.

I haven't figured out how to get the user to grant this permission, and currently the following request

https://graph.facebook.com/userid/feed?access_token=123|abc-def&limit=4&locale=en

results in error 404 (as one would expect).

There is a lot of (outdated) tutorials about how to create an app to pull a user feed from Facebook, but none of them tell you how to get the users_posts permission from the user that owns the feed. It looks like you didn't need this before, but Facebook has recently tightened their privacy protection and the old ways no longer work.

How can users grant my app the users_posts permission?

(I am using my own user to test this, so this is not about coaxing the user to grant the permission - it is about how you code the app so that the user actually can do this.)

Answer 1

I suggest using the JS SDK for login, because it is a lot easier. For example, this is how you authorize users with the user_posts permission:

FB.login(function(response) {
    if (response.authResponse) {
        //user just authorized your app
        console.log(response);
    }
}, {scope: 'user_posts', return_scopes: true});

After that, you can just use this to get the posts:

FB.api('/me/feed', {limit: 4}, function(response) {
    console.log(response);
});

Source: http://www.devils-heaven.com/facebook-javascript-sdk-login/

---

If you want to login with the PHP SDK, read this: https://developers.facebook.com/docs/php/howto/example_facebook_login

The relevant part from the docs:

$permissions = ['user_posts'];
$loginUrl = $helper->getLoginUrl('https://example.com/fb-callback.php', $permissions);