This question already has an answer here:
I am using the below PHP code but it is generating an error/warning message:
PHP Notice: Undefined offset: 1
The code runs as expected.
while(1) {
while($data = fgets($irc, 128)) {
$ex = explode(' ', $data);
if($ex[0] == "PING"){
fputs($irc, "PONG ".$ex[1]."
");
}
if($ex[1] == "265"){
$lusers = str_replace(',', '', $ex[8]);
$gusers = $ex[10];
}
}
}
I cannot see what is wrong in the code. It is running as expected so why is the error/warning message being generated?
Suggestion to use print_r($ex) shows:
Array
(
[0] => SeIKVfMCuzNTGjZ
)
PHP Notice: Undefined offset: 1
Should I just add a check that $ex > 0 or is there a better way?
I updated code so it only runs if $ex > 0:
while(1) {
while($data = fgets($irc, 128)) {
$ex = explode(' ', $data);
if($ex[0] == "PING"){
fputs($irc, "PONG ".$ex[1]."
");
}
if(count($ex) > 0){
if($ex[1] == "265"){
$lusers = str_replace(',', '', $ex[8]);
$gusers = $ex[10];
}
}
}
}
?>
but I still get the error/warning message. How is it possible?
</div>
This should work for you:
if(isset($ex[1]) && $ex[1] == "265") {
//do stuff
} else {
echo "Index doesn't exist!";
}