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Date:Wed Aug 20 19:12:33 2014
Subject:Re: Re: socket gives notice on server side(simple newbie code)
Sincerely
Negin Nickparsa


On Wed, Aug 20, 2014 at 9:05 AM, Robert Williams <[email protected]>
wrote:

> I suspect there's a line-feed character in the input value. Try changing
> this:
>
> if($stock[$input]!=NULL)
>
> to this:
>
> if($stock[trim($input)]!=NULL)
>
> --
> Robert E. Williams, Jr.
> Senior Vice President of Software Development
> Newtek Businesss Services, Inc. -- The Small Business Authority
> https://www.newtekreferrals.com/rewjr
> http://www.thesba.com/
>
>
>
​Robert actually you were right! var_dump had the string(7) "salam" so it
is not string(4) and that is the problem the problem was somewhere else
which I didn't get it at first
thank you guys! great help!

I learned that I should trim my string even when it is string with STDIN it
would include the quotes in the string too​



> On Aug 20, 2014, at 08:09, Jim Giner <[email protected]> wrote:
>
> > 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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> >
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