Re: socket gives notice on server side(simple newbie code)

Negin Nickparsa wrote: > I am struggling with my code and I cannot figure out what I have done which > it tells me Notice > > here is my server.php: > > <?php > set_time_limit(0); > > $stock=array("hi"=>"hi","name"=>"negin","question"=>"answer"); > $address="127.0.0.1"; > $port="127"; > $sock= socket_create(AF_INET,SOCK_STREAM, getprotobyname("tcp")); > socket_bind($sock, $address,$port); > socket_listen($sock); > $client= socket_accept($sock); > $input=socket_read($client, 1024); > echo "you entered ".$input; > if($stock[$input]!=NULL) > { > $answer=$stock[$input]; > socket_write($sock,$answer, $port); > > } > else{ > echo socket_strerror(0); > } > socket_close($sock); > ?> > > and here is my client.php: > > <?php > $address="127.0.0.1"; > $port="127"; > $sock= socket_create(AF_INET,SOCK_STREAM, getprotobyname("tcp")); > echo "enter your data:"; > $input= fgets(STDIN); > socket_connect($sock, $address,$port); > socket_write($sock,$input ,strlen($input)); > $dt=socket_read($sock, 1024); > echo $dt; > ?> > > I really cannot undrestand why the output is this on server side: > > c:\wamp\www\phpzend>php -f server.php > you entered salam > > Notice: Undefined index: salam > in C:\wamp\www\phpzend\server.php on line 12 > > Call Stack: > 0.0010 232328 1. {main}() C:\wamp\www\phpzend\server.php:0 > > The operation completed successfully. > > but it doesn't write on my client side so this notice completely is a big > issue I var_dumped the $input and it is string I have no idea what's wrong > here Have you checked that var_dump($input) prints string(5) "salam" (note the string length 5). -- Christoph M. Becker