php variable in query - postgresql

Question

I have problem with query value.

This is working.

$userArray = $database->select("app", "fb_id=14175962160");

but below isn't working. Why?

$user = $facebook->getUser(); // contains my fb id - 14175962160

$userArray = $database->select("app", "fb_id='$user'");
$userArray = $database->select("app", "fb_id=$user");
$userArray = $database->select("app", "fb_id=" . $user);

What do you mean with "isn't working"? What error message(s) do you get? — Markus Deibel
– Markus Deibel, Commented Jun 1, 2013 at 21:33
Are you trying to run this on a 32bit server? You may be experiencing integer overflow. stackoverflow.com/questions/6505203/… Did you var_dump($user) to see what value it actually contains? — Michael Berkowski
– Michael Berkowski, Commented Jun 1, 2013 at 21:34
I don't believe, that $user = 14175962160, because in this case "fb_id=$user" string is the same as the working string. — user4035
– user4035, Commented Jun 1, 2013 at 21:36
ah the $user is int(1417596216) and should be int(14175962160). But I don't know why the zero is missing. (string) $user == string(10) "1417596216" — dfgd
– dfgd, Commented Jun 1, 2013 at 21:46
@VladimírVolek Try to print $facebook->getUser(); instead of saving it into the variable. What does it give? — user4035
– user4035, Commented Jun 1, 2013 at 22:09

dfgd · Accepted Answer · 2013-09-04 09:41:14Z

0

The whole problem was that I manually saved bad fb id to the database (With zero on the end) and I didn't notice that for a day. that's all :D. Thank you guys

edited Sep 4, 2013 at 9:41

answered Jun 3, 2013 at 9:18

dfgd

4551 gold badge5 silver badges20 bronze badges

Add a comment |

Collectives™ on Stack Overflow

php variable in query - postgresql

1 Answer 1

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Your Answer

Sign up or log in

Post as a guest

Linked

Related