Handling big user IDs returned by FQL in PHP

问题:

I'm using FQL to retrieve a list of users from Facebook. For consistency I get the result as JSON. This causes a problem - since the returned JSON encodes the user IDs as numbers, json_decode() converts these numbers to floating point values, because some are too big to fit in an int; of course, I need these IDs as strings.

Since json_decode() does its own thing without accepting any behavior flags, I'm at a loss. Any suggestions on how to resolve this?

回答1:

json_decode() can convert large integers to strings, if you specify a flag in the function call:

$array = json_decode($json, true, 512, JSON_BIGINT_AS_STRING)

回答2:

I've resolved the issue by adding &format=json-strings to my the FQL api call, like so:

$myQuery = "SELECT uid2 FROM friend WHERE uid1=me()";
$facebook->api("/fql?q=" . urlencode($myQuery) . "&format=json-strings")

This tells facebook to wrap all the numbers in quotes, which leads json_decode to use neither int-s not floats.

Because I was afraid this issue is not restricted to FQL but to all graph API calls that choose to represent some of the IDs as BIG-INTs I've went as far as patching facebook's PHP SDK a bit to force Facebook to return all of its numbers as strings.

I've added this one line to the _graph function. This would be line 738 in facebook_base.php, version 3.1.1

$params['format'] = 'json-strings';

Sure fix

回答3:

I had a similar problem where json_decode was converting recent twitter/tweet IDs into exponential numbers.

Björn's answer is great if you want your BIGINT to become a string - and have PHP 5.3+. If neither of those things are true, another option is to up PHP's float precision. This can be done a different few ways...

• find the precision value in your php.ini and change it to precision = 20
• add ini_set('precision', 20); to your PHP app
• add php_value precision 20 to your app's .htaccess or virtual host file

回答4:

Quick and dirty, seems to work for now :

$sJSON = preg_replace('/:(\d+)/', ':"${1}"', $sJSON);

回答5:

I use this and it works almost great.

json_decode(preg_replace('/("\w+"):(\d+)/', '\\1:"\\2"', $jsonString), true)

The json breaks when there is geo data included, eg. {"lat":54.2341} results in "lat":"54".2341

Solution:

$json = preg_replace('/("\w+"):(\d+)(.\d+)?/', '\\1:"\\2\\3"', $json);

回答6:

This (preg_replace('/("\w+"):(\d+)(.\d+)?/', '\\1:"\\2\\3"', $json);) worked for me (for parsing result from facebook api)