Question

I wrote the following FQL query which should work but it seems to fail:

SELECT coords, tagged_uids, page_id FROM checkin 
WHERE author_uid = IN (SELECT uid2 FROM friend WHERE uid1 = me())

It throws an error as follows:

Parser error: unexpected ‘IN’ at position 68.

Doesn’t make any sense when other queries of similar syntax work fine like this one:

SELECT uid, name, pic_square FROM user 
WHERE uid IN (SELECT uid2 FROM friend WHERE uid1 = me())

Anyone know a better way to get the check-ins from my friends?

Answer 1

IN should replace the = operator, and you only need one. Try:

SELECT coords, tagged_uids, page_id FROM checkin 
WHERE author_uid IN (SELECT uid2 FROM friend WHERE uid1 = me())

Also, make sure you have the friends_checkins extended permission.