Richard,
No, the code was cut from the php file and displayed as is in my e-mail. I
also tried to see if I could declare the two suspect variables in another
IF() statement before the IF() statement that's fouling the execution of the
.jpg construct, but no luck there too. Could I just be overburdening the
program? I've got a lot of code (about 150 lines) in the .jpg construct.
Cheers,
Hugh
----- Original Message -----
From: Richard Lynch <[EMAIL PROTECTED]>
To: Hugh Danaher <[EMAIL PROTECTED]>
Sent: Wednesday, August 22, 2001 12:18 AM
Subject: Re: [PHP] Re: if statement and imageline()
> Hmmmmmm.
>
> Are you sure you didn't have:
>
> if ($year = $startyear)
>
> *THAT* would assign 1996 over and over to $year, after the ++, so it would
> never move forward, and the script would time out...
>
> Not sure what else could be wrong, if it's *JUST* the if part you are
> adding...
>
> --
> WARNING [EMAIL PROTECTED] address is an endangered species -- Use
> [EMAIL PROTECTED]
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> ----- Original Message -----
> From: Hugh Danaher <[EMAIL PROTECTED]>
> To: Richard Lynch <[EMAIL PROTECTED]>
> Cc: Php-General <[EMAIL PROTECTED]>
> Sent: Wednesday, August 22, 2001 1:05 AM
> Subject: Re: [PHP] Re: if statement and imageline()
>
>
> > Richard,
> >
> > Without the code below, I get a good .jpg file and it displays nicely.
> But,
> > with the IF statement in, I get a time out error message "Execute time
> > exceeds 30 seconds..." I think it is because two of the variables are
> > declared after the IF statement, even though they aren't being used on
the
> > first loop. I'll try to declare them before the FOR loop and see if
that
> > works.
> >
> > As you might tell from my code, the only software course I ever had, was
> in
> > college 20 years ago, and it was in BASIC language (with line numbers!).
> > However, I find php easy to work with, and have done some amazing things
> > with it in the last 2 months. With your help and the help of others,
I'll
> > certainly make progress.
> >
> > Thanks,
> > Hugh
> > ----- Original Message -----
> > From: Richard Lynch <[EMAIL PROTECTED]>
> > To: <[EMAIL PROTECTED]>
> > Sent: Tuesday, August 21, 2001 5:16 PM
> > Subject: [PHP] Re: if statement and imageline()
> >
> >
> > > What are you getting? An image at all? Broken image?
> > >
> > > Try getting rid of the header("image/jpeg") or whatever, and see if
you
> > have
> > > some error message.
> > >
> > > --
> > > WARNING [EMAIL PROTECTED] address is an endangered species -- Use
> > > [EMAIL PROTECTED]
> > > Wanna help me out? Like Music? Buy a CD:
http://l-i-e.com/artists.htm
> > > Volunteer a little time: http://chatmusic.com/volunteer.htm
> > > ----- Original Message -----
> > > From: Hugh Danaher <[EMAIL PROTECTED]>
> > > Newsgroups: php.general
> > > To: Php-General <[EMAIL PROTECTED]>
> > > Sent: Tuesday, August 21, 2001 1:16 AM
> > > Subject: if statement and imageline()
> > >
> > >
> > > help,
> > >
> > > I am trying to set up a .jpg file to graph the earnings per year of a
> > > company. I can generate a log-normal graph, can get it to display the
> > > earnings per year as circles on the graph, but I can't get the " if
()
> "
> > > statement to work. I know I am setting two of the variables in the if
> > > statement after the if statement executes once, but these variables
> won't
> > be
> > > used until after the " for () " loops once, and therefore should be
> > > available for use in " imageline() " on the second loop (where
> > > $year>$startyear). Somehow, I think my logic is correct but it
mustn't
> be
> > > so.
> > >
> > >
> > > $startyear=1996;
> > > $chart_start_year=1992;
> > > for ($year=$startyear;$year<=$startyear+7;$year++)
> > > {
> > > $x=(($year-$chart_start_year)*20)+20;
> > > $y=420-log(${"earnings_".$year})*75;
> > > if ($year>$startyear)
> > > {
> > > imageline($image,$first_x,$first_y,$x,$y,$blue);
> > > }
> > > $first_x=$x;
> > > $first_y=$y;
> > >
> > imagettftext($image,9,0,$x_distance-4,$y_distance+3,$blue,$font2,"m");
> > > }
> > >
> > >
> > >
> > > --
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> > >
>
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