How to get all the Open Graph Beta actions generated by an app?
Asked Answered
B

1

15

It is already possible to get all actions for custom app actions and objects:

https://graph.facebook.com/me/{appNameSpace}:{action}/{object}

This will list all the actions generated by an app, but only for a given user...

How do I get all the actions generated by an app (for all its users)? I've tried this request with an app access token :

https://graph.facebook.com/{appId}/{appNameSpace}:{action}/{object}

But it does not work...

I was also looking for an activities or actions FQL table, since each action has an id, they should be accessible somewhere.

Any suggestions?

Bruit answered 4/11, 2011 at 10:17 Comment(0)
S
4

AFAIK there is no way to get all actions published by application via OpenGraph, but your application creates those actions, why just not record results of this operation?

Then you create action you should get response like this:

{
  id: “{action-instance-id}”
}

Later you may read this action:

GET https://graph.facebook.com/{action-instance-id}

You for sure may benefit from doing request to get multiple actions like this:

GET https://graph.facebook.com/?ids={action-id1},{action-id2},{action-idn}

And even batch those requests

Sanburn answered 29/12, 2011 at 20:51 Comment(8)
how do apps like Washington Post Social Reader display all of your friends' actions?Chittagong
They store them on their end.Highwrought
Or they iterate across your friends and get all the actions from all friends' graphs.Ozzy
Seems like iterating across all friends would take too long since there is no FQL available to do this (see #8499797). How would you implement it?Chittagong
You still can batch up to 50 requests which can help do it much fasterSanburn
Is possible to get a list of friends using the app? maybe that diminish the number of friends to ask for ActionsBaste
@gonxalo, it is possible, just search it on Stack Overflow and you'll get a bunch of questions with answer to this question...Sanburn
i know. it was rhetorical my question. :)Baste

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