Re: [PHP] Re: if statement and imageline()
Richard, I let this problem sit for awhile, did some other work, and looked at it again. This time, I printed out all the values and found that some values =INF. No wonder it died. Since I'm making a log-normal chart, occasionally I'm trying to calculate the log of zero--an infinitely small number which can't be calculated! I did try the error_log statement but my ISP puked on the syntax. - Original Message - From: Richard Lynch [EMAIL PROTECTED] To: Hugh Danaher [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 1:21 PM Subject: Re: [PHP] Re: if statement and imageline() No. 150 lines is chump change. See if you can get ImageLine to work in a simpler fashion. Also, try adding: error_log(ImageLine: $start_x $start_y $x $y, E_NOTICE); And then look at your Apache logs. You'll at least be able to see if the program is getting there while it tries to draw stuff. -- WARNING [EMAIL PROTECTED] address is an endangered species -- Use [EMAIL PROTECTED] Wanna help me out? Like Music? Buy a CD: http://l-i-e.com/artists.htm Volunteer a little time: http://chatmusic.com/volunteer.htm - Original Message - From: Hugh Danaher [EMAIL PROTECTED] To: Richard Lynch [EMAIL PROTECTED] Cc: Php-General [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 3:50 AM Subject: Re: [PHP] Re: if statement and imageline() Richard, No, the code was cut from the php file and displayed as is in my e-mail. I also tried to see if I could declare the two suspect variables in another IF() statement before the IF() statement that's fouling the execution of the .jpg construct, but no luck there too. Could I just be overburdening the program? I've got a lot of code (about 150 lines) in the .jpg construct. Cheers, Hugh - Original Message - From: Richard Lynch [EMAIL PROTECTED] To: Hugh Danaher [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 12:18 AM Subject: Re: [PHP] Re: if statement and imageline() Hmm. Are you sure you didn't have: if ($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] 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] 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
Re: [PHP] Re: if statement and imageline()
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() Hmm. 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] 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] 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); } -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Re: if statement and imageline()
Richard, Thanks for your help. I'll try the error_log() and see what comes out. Hugh - Original Message - From: Richard Lynch [EMAIL PROTECTED] To: Hugh Danaher [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 1:21 PM Subject: Re: [PHP] Re: if statement and imageline() No. 150 lines is chump change. See if you can get ImageLine to work in a simpler fashion. Also, try adding: error_log(ImageLine: $start_x $start_y $x $y, E_NOTICE); And then look at your Apache logs. You'll at least be able to see if the program is getting there while it tries to draw stuff. -- WARNING [EMAIL PROTECTED] address is an endangered species -- Use [EMAIL PROTECTED] Wanna help me out? Like Music? Buy a CD: http://l-i-e.com/artists.htm Volunteer a little time: http://chatmusic.com/volunteer.htm - Original Message - From: Hugh Danaher [EMAIL PROTECTED] To: Richard Lynch [EMAIL PROTECTED] Cc: Php-General [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 3:50 AM Subject: Re: [PHP] Re: if statement and imageline() Richard, No, the code was cut from the php file and displayed as is in my e-mail. I also tried to see if I could declare the two suspect variables in another IF() statement before the IF() statement that's fouling the execution of the .jpg construct, but no luck there too. Could I just be overburdening the program? I've got a lot of code (about 150 lines) in the .jpg construct. Cheers, Hugh - Original Message - From: Richard Lynch [EMAIL PROTECTED] To: Hugh Danaher [EMAIL PROTECTED] Sent: Wednesday, August 22, 2001 12:18 AM Subject: Re: [PHP] Re: if statement and imageline() Hmm. Are you sure you didn't have: if ($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] 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] 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_
[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); } -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
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); } -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]