PHP 'fwrite' Will Not Take Variable

Question

I'm creating an IRC bot, and I wanted to store some functions in a different file. I reference the file in my main file with include() and it works, but when I try to give fwrite() a variable like so:

$data ="THIS IS DATA\r\n";
fwrite($sock, $data);

It throws this error:

PHP Warning: feof() expects parameter 1 to be resource, null given in /home/kinz/Desktop/Eve/Eve.php on line 18

PHP Warning: fgets() expects parameter 1 to be resource, null given in /home/kinz/Desktop/Eve/Eve.php on

But when I put:

fwrite($sock, "THIS IS DATA\r\n");

It works perfectly fine. Here is my full code:

function connect_s ($serv, $port, $nick) { $sock = fsockopen($serv, $port); fwrite($sock, "USER $nick $nick $nick :$nick\r\n"); fwrite($socket, "NICK $nick\r\n"); }
function write_d ($data) { fwrite($sock, $data); }
function join_c ($chan) { write_d("JOIN $chan\r\n"); }
function get_in ($in) { for ($i = 3; $i <= count($in); $i++) { $out .= " $in[$i]"; } return trim("$out\r\n"); }
function change_n ($nick) { write_d("NICK $nick\r\n"); } ?>

I want to know why I can't use a variable as the second parameter in fwrite. Any help is greatly appreciated

And yet when I set parameter 2 manually, it fixes the problem.. — Kinz
– Kinz, Commented Mar 18, 2012 at 1:11
$sock is not resource id, make sure that u open server and assign to $sock ($sock = fsockopen...) before — safarov
– safarov, Commented Mar 18, 2012 at 1:11
Try var_dump($sock). Just because setting the second parameter removes the warning doesn't mean it worked. — stan
– stan, Commented Mar 18, 2012 at 1:12

Jason · Accepted Answer · 2018-02-13 14:39:20Z

5

Each new function has it's own scope for the variables. You've defined $sock in one function but it's not defined in the other functions. You need to do one of 3 things:

Return $sock and pass it to any subsequent functions
have global $sock; at the beginning of each of your functions
write this all in an object class, and assign and use $this->sock as opposed to $sock

3 would be the "best" solution, 1 would be how to accomplish this in a functional context, and 2 is probably the easiest but using globals in this way is the last resort.

edited Feb 13, 2018 at 14:39

answered Mar 18, 2012 at 1:12

Jason

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Comments

bluegman991 · Accepted Answer · 2012-03-18 01:13:00Z

1

$sock is not defined in your function named write_d.

You need to make $sock a global variable or put all of these functions into a class and make $sock a property of the class.

answered Mar 18, 2012 at 1:13

bluegman991

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