What steps will reproduce the problem?
1. use this code;
$t = new Twitter();
$t->username='un';
$t->password='pw';
$r = $t->userTimeline(false, 1);
print_r($r);
2. try and access any of object properties... they all have a property name
that is a number so you can't
What is the expected output? What do you see instead?
expected an array, got an object
What version of the product are you using? On what operating system?
1.0beta, linux, PHP 5.2.6
Please provide any additional information below.
Line 466 reads "return (object) json_decode( $data );"
if the JSON is an array it returns an array! You then turn it into an
object and this makes no sense... there is no need for typecasting here.
if its an object the result ($r as above) would need to be accessed like
this;
print_r($r->1);
but thats impossible.
if you take out "(object)" the following code then works fine.
print_r($r[0]);
Original issue reported on code.google.com by douga...@gmail.com on 20 Oct 2008 at 9:03
Original issue reported on code.google.com by
douga...@gmail.com
on 20 Oct 2008 at 9:03