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imagecreatefrompng() times out


paymentstv

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Hello All,

 

I have tried to find a solution for this issue for days. The script was working fine until yesterday when all of a sudden I get this warning.

 

Warning: imagecreatefrompng(http://domain.com/images/broadcast1.png) [function.imagecreatefrompng]: failed to open stream: Connection timed out in /home/domain/public_html/logo/logo1.php on line 8

 

line 8 is:

$image_1 = imagecreatefrompng($file_1);

Where

$file_1 = "http://domain.com/images/broadcast1.png";

 

The exact code is working perfectly fine in a different server I have, both servers are running Centos with PHP5 + GD fully enabled. The image http://domain.com/images/broadcast1.png (~5Kb) is loading very fast when open through a browser.

 

Please please help me resolve this, I have tried many different things and even uninstall and install GD again but n luck. Thank you so much for your help.

 

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https://forums.phpfreaks.com/topic/221574-imagecreatefrompng-times-out/
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Oh you might wanna set:

set_time_limit(60);

So it doesn't time out...

 

And you may also wanna try:

$filecontent=file_get_contents('http://www.phpfreaks.com');

for example from phpfreaks, to see if it can load stuff from other sites...

 

try to load other images that is located on a network.

 

It would also be nice if you gave us the exact image url.

Oh you might wanna set:

set_time_limit(60);

So it doesn't time out...

 

And you may also wanna try:

$filecontent=file_get_contents('http://www.phpfreaks.com');

for example from phpfreaks, to see if it can load stuff from other sites...

 

try to load other images that is located on a network.

 

It would also be nice if you gave us the exact image url.

 

1. The timeout occurs at about 90Seconds now but since it is a very active site that is not a solution

2. I have tried another image http://www.drweb.de/icons/twitter/pd_twitter_iconset/pd_twitter_iconset/PNG/256/icon_c.png now I get a  different error

 

Warning: imagecreatefromjpeg(http://domain.com/logo/logo.php?txt=By:cnn.com) [function.imagecreatefromjpeg]: failed to open stream: Connection timed out in /home/domain/public_html/logo/logo1.php on line 15

 

Where line 15 is

$image_2 = imagecreatefromjpeg($file_2);

Where;

$image_2= "http://domain.com/logo/logo.php?txt=By:$name";

//This logo.php converts the string By:$name to a jpeg file. logo.pho is working as I can see the image when I try this in a browser.

 

3.I am sorry because I am not authorized to put the whole code or site name in public forums. Please accept my apologies.

Since the code worked 100% until tomorrow morning I believe it has to do with php installation/ WHM/Cpanel/ GD..etc but I need help to figure out where I should look.

 

 

When I tried to see that picture:

Forbidden

You don't have permission to access /icons/twitter/pd_twitter_iconset/pd_twitter_iconset/PNG/256/icon_c.png on this server.

Additionally, a 302 Found error was encountered while trying to use an ErrorDocument to handle the request.
Apache/2.2.3 (Debian) PHP/5.2.8-0.dotdeb.1 with Suhosin-Patch mod_ssl/2.2.3 OpenSSL/0.9.8c Server at www.drweb.de Port 80

 

Besides:

$image_2= "http://domain.com/logo/logo.php?txt=By:$name";

$image_2 = imagecreatefromjpeg($file_2);

$image_2!=$file_2

May be you did a typo here now though, and it might be the installation of something around GD...

I was able to see the image fine but I will use another png file hosted in a spare server.

http://173.244.194.170:805/item.png

 

When I use that image in;

$file_1 = "http://173.244.194.170:805/item.png";

$image_1 = imagecreatefrompng($file_1);

 

I get the original error back.

Warning: imagecreatefrompng(http://173.244.194.170:805/item.png) [function.imagecreatefrompng]: failed to open stream: Connection timed out in /home/veedo/public_html/logo/logo1.php on line 8

 

I have a feeling this might have something to do with URL open ability of php?

 

tried this code

<p>PHP version is: <?php echo phpversion(); ?></p>

<p>Safe mode is: <?php echo ($bSafeMode) ? "On (this is usually bad)" : "Off (this is good!)"; ?></p>

<?php
    /*
     * Try opening www.microsoft.com
     */
@$fIn = fopen("http://www.domain.com/logo/logo.php?txt=By:cn.com","r");
if ($fIn !== FALSE)
{
    echo "<p>URL open succeeded!</p>";
fclose($fIn);
}
else{
echo "<p>URL open failed: $php_errormsg</p>";}

$filecontent=file_get_contents('http://www.phpfreaks.com');
echo $filecontent;
?>

 

Result:

PHP version is: 5.2.14

Safe mode is: Off (this is good!)

URL open succeeded!

Subscribe to PHP Freaks RSS

And after that I see contents of phpfreaks site.

 

When I try the local file it display the image correctly.

However, second part of the image is not created

 

$imageurl = "http://domain.com/logo/logo.php?txt=By:";
$file_2 = $imageurl.$sitename;//$sitename is a variable passed on to this logo1.php
$image_2 = imagecreatefromjpeg($file_2);

 

I create two images image_1 and image_2 and then use php to copy them to a single image.

In this case only the first image (local png file) is displayed.

No errors reported.

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