How can I get the Facebook identifiers of all of my Facebook friends? I mean the numbers like 7843652532 or 93923423.

Either via an URL, a Facebook app, a browser extension, or anything.

Any format would be OK, for instance a text list or JSON.

(I am using these IDs to associate with my address book, since many of my friends change names.)

  • @Barry: Yes :-) Any format would be OK, for instance list or JSON. – nic Feb 2 '11 at 12:10

You get this is JSON format by going to http://developers.facebook.com/docs/reference/api

About the 4th paragraph you need to click the link next to "Friends" (which is https://graph.facebook.com/me/friends) however the link on the page includes an Authorization token which is require to run the query.

Facebook Friends Link

This will give a JSON result of all friends names and Facebook ID's

  • Impressive! Not need to write a request or authorize an application or anything :-) – nic Feb 2 '11 at 12:49
  • The first link displays a page which has probably changed since 2011, now it does not have a "Friends" section. Would you mind updating? Thanks! – nic Jan 17 '19 at 7:13

Since December 2014 it is a bit complex:

  • Install the FB Purity extension
  • Load the Facebook homepage
  • Find your ~/.config/chromium/Default/Local Storage/chrome-extension_ncdlagniojmheiklojdcpdaeepochckl_0.localstorage file. (exact name might vary, search for ncdlagniojmheiklojdcpdaeepochckl in your filesystem)
  • Open this file with SQLitebrowser
  • Select the data ItemTable>oldfriendstore-<something>
  • It is the list of ID,name couples as JSON

PRE-2014/12 method

Barry's answer has worked great in the past, but that link is not available anymore. New solution: Go to https://developers.facebook.com/tools/explorer/?method=GET&path=me/friends select V1.0 and press Submit. You get the whole list of ID+name couples.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.