PHP - Parsing Source

by any means is there a way to view a email.php form source? not just the html part but the entire script including the php?

This is source of a page:
http://vn-ace.com/forum/test.txt


I want to get contents betweent
<blockquote class="postcontent restore">()
and
</blockquote>()
but i dont know how to do ?
How can i do ? Please help me. I am newbie.
Thanks

when i right click to view page source i just saw the javascript calling php as source file , while on page there is table,
i need to read that table in database , is there nay way to do that, file_get contents dont work on it its just take javascript not the page table

So I am completely done with my forum after several posts here and a lot of time. But crap! I just realized that the way my avatar system works it will give away the password! I REALLY don't want to redo that system because truth is it is about 40 percent of the entire sites coding. It works by making pictures in a directory named the usersname.thepassword . Whatever filesystem. Now when I echo the path everyone can see the password and username in the source code! And thy can click it to see the picture! Is there a way to hide the paths or the source codeM thanks!

Hey folks,   I'm looking for an open-source php project to contribute to. I'm not looking for a really well-known project (ex: Wordpress, Zencart) Something small would be greatly appreciated!

So I am completely done with my forum after several posts here and a lot of time. But crap! I just realized that the way my avatar system works it will give away the password! I REALLY don't want to redo that system because truth is it is about 40 percent of the entire sites coding. It works by making pictures in a directory named the usersname.thepassword . Whatever filesystem. Now when I echo the path everyone can see the password and username in the source code! And thy can click it to see the picture! Is there a way to hide the paths or the source codeM thanks!

How to edit my source code so the output prints correctly ? 
Check the last line in screenshot. 

 

i am currently creating a school portal. there is this page which enables user to create their own website for Computer Education lesson. Every user are able to upload their files and view but the problem is, i need a function for the user to be able to edit it in on the page itself.

Hi Everyone, I am a noob so forgive me I tried looking this up myself But I am still confused.

Basically, I am trying to use the above code to display a gallery of images from my images folder on my image site.

But this free code I found below Pulls the images from the /root not a named directory where the images are stored.

Would someone be kind enough to tell me how I can get the script below to pull the large images from a folder called "images" Vs the script looking for the images the root.

Thanks so much in advance 

Code: [Select]
<?php
# SETTINGS
$max_width = 100;
$max_height = 100;

function getPictureType($ext) {
if ( preg_match('/jpg|jpeg/i', $ext) ) {
return 'jpg';
} else if ( preg_match('/png/i', $ext) ) {
return 'png';
} else if ( preg_match('/gif/i', $ext) ) {
return 'gif';
} else {
return '';
}
}

function getPictures() {
global $max_width, $max_height;
if ( $handle = opendir(".") ) {
$lightbox = rand();
echo '<ul id="pictures">';
while ( ($file = readdir($handle)) !== false ) {
if ( !is_dir($file) ) {
$split = explode('.', $file); 
$ext = $split[count($split) - 1];
if ( ($type = getPictureType($ext)) == '' ) {
continue;
}
if ( ! is_dir('thumbs') ) {
mkdir('thumbs');
}
if ( ! file_exists('thumbs/'.$file) ) {
if ( $type == 'jpg' ) {
$src = imagecreatefromjpeg($file);
} else if ( $type == 'png' ) {
$src = imagecreatefrompng($file);
} else if ( $type == 'gif' ) {
$src = imagecreatefromgif($file);
}
if ( ($oldW = imagesx($src)) < ($oldH = imagesy($src)) ) {
$newW = $oldW * ($max_width / $oldH);
$newH = $max_height;
} else {
$newW = $max_width;
$newH = $oldH * ($max_height / $oldW);
}
$new = imagecreatetruecolor($newW, $newH);
imagecopyresampled($new, $src, 0, 0, 0, 0, $newW, $newH, $oldW, $oldH);
if ( $type == 'jpg' ) {
imagejpeg($new, 'thumbs/'.$file);
} else if ( $type == 'png' ) {
imagepng($new, 'thumbs/'.$file);
} else if ( $type == 'gif' ) {
imagegif($new, 'thumbs/'.$file);
}
imagedestroy($new);
imagedestroy($src);
}
echo '<li><a href="'.$file.'" rel="lightbox['.$lightbox.']">';
echo '<img src="thumbs/'.$file.'" alt="" />';
echo '</a></li>';
}
}
echo '</ul>';
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" dir="ltr" lang="en-US">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UFT-8" />
<title>Pictures</title>
<link rel="stylesheet" href="css/lightbox.css" type="text/css" media="screen" />
<style type="text/css">
#pictures li {
float:left;
height:<?php echo ($max_height + 10); ?>px;
list-style:none outside;
width:<?php echo ($max_width + 10); ?>px;
text-align:center;
}
img {
border:0;
outline:none;
}
</style>
</head>
<body>

<?php getPictures(); ?>


<script type="text/javascript" src="js/prototype.js"></script>
<script type="text/javascript" src="js/scriptaculous.js?load=effects,builder"></script>
<script type="text/javascript" src="js/lightbox.js"></script>
</body>
</html>

Hello,

Where can i find the source of the pre made php functions?

Greets,
Scuar.

Hello all,

i have a feeling im doing something wrong, but i have no idea what.

being server-side code, php should not show when you 'view page source' in your browser (or rather it should display as html), correct ?

why, then am i seeing php code when i view page source?

see attached image

how to copy a full folder from source to destination using php

please help me.

As part of the registration procedure, my PHP application generates the mail below and presents the option to a user to click on the url in the mail to activate his account.

Please click on this link
http://www.example.org.ng/activate.php?token=XeZNYf8uDVYxAY5+RBqldOosI1hm/FjB0cLnXB8R
to activate your account.

The activate.php script returns that their is no record of this token in the database, even though it is there. In troubleshooting, i printed the $token = $_GET["token"] in the activate.php script; and this is what i got

XeZNYf8uDVYxAY5 RBqldOosI1hm/FjB0cLnXB8R

.   Notice that the $token variable is missing one character, (the +), which is the 16th character form the left!!   Why this would happen is unclear. Any thoughts.   If it helps, the is the script generating my random tokens:

function generateToken($length = 40)
{
   if(function_exists('openssl_random_pseudo_bytes')) 
   {
     $token = base64_encode(openssl_random_pseudo_bytes($length, $strong));
     if($strong == TRUE)
     return substr($token, 0, $length); //base64 is about 33% longer, so we need to truncate the result
   }
  //fallback to mt_rand if php < 5.3 or no openssl available
  $characters = '0123456789';
  $characters .= 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz^!$'; 
  $charactersLength = strlen($characters)-1;
  $token = '';
  //select some random characters
  for ($i = 0; $i < $length; $i++) 
  {
    $token .= $characters[mt_rand(0, $charactersLength)];
  }        
  return $token;
}
$token=generateToken($length = 40);

Edited by terungwa, 09 June 2014 - 05:17 PM.

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=355719.0

I've got another funny one...

I've noticed that sometimes on some of my pages (all of which are generated by PHP), when i click view source, the source it shows me is not from the page i'm looking at. This isn't a problem that affects me in any way (not yet anyway), it's just a bit odd.

I've noticed it happens a little more often in chrome than any other browser.

Has anyone else noticed anything like this before? I'm intrigued to know the cause.

The only real google response was this: http://www.mail-archive.com/debian-bugs-dist@lists.debian.org/msg144609.html and i haven't seen another issue similar on here.