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

[Jim Giner <jim.giner@xxxxxxxxxxxxxxxxxx>]

On 8/20/2014 11:00 AM, Negin Nickparsa wrote:
> echo "you entered ".$input;
> if($stock[$input]!=NULL)
> {
>      $answer=$stock[$input];
>      socket_write($sock,$answer, $port);
>
> }
> else{
>      echo socket_strerror(0);
> }

How about adding the following before the above lines:
if (!isset($stock[$input]))
   print_r($stock);

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