PHP - Open File In Another Language

Hi guys

<?php
sleep(00);
$handle = fopen("http://www.myurl.com/?bla=bla=bla" , "r");
fclose($handle);

?>

How do i prevent this from outputing to a file on my server. I only need the url to be accessed to add a new row of data to my DB.

Any Ideas?
Regards
Mark

Hello all, i am go9090go.
Today i made a domains for a jar file people can upload from my website.
I made this to make the jar file close source and its easy to update.
Now i made a java classloader and everything i made works.
The classloader call a php document with the password and username.
The pass and name will be checked inside a databse and if its inside i use
header() to load the jar file.

But when i just go to my main domain i get the index of the site and people can easly download the jar file without have to walk thru the php pass checker.
So i want to place the jar file inside a protected folder,and i want that only way you get acces to this jar is by the php file. How can i get a file from a protected folder?

here is the php used when the jar file is not inside a protected folder:

<?php

$DBName = "name";//name database        
$DBUser = "name";//user        
$DBPassword = "pass"; //passs
$DBHost = "host"; //might be different
         
mysql_connect($DBHost, $DBUser, $DBPassword);
        mysql_select_db($DBName);

        $username = $_GET['username'];
        $password = $_GET['password'];

$IP = $_SERVER['REMOTE_ADDR'];
    
$string = "Java";
$pos = strpos($agent, $string);    
if (!strpos($_SERVER['HTTP_USER_AGENT'], "Java")) 
{        
echo("Your Auth has been banned for trying to breach security.");        
//mysql_query("delete from users where username='$username'");
exit();    
}

$query = "select * from users where name='$username' and pass='$password'";
        mysql_query($query);
        $num = mysql_affected_rows();
if ($num > 0) 
{
header('Location:script/Script.jar');
}

?>


now i want to use the header to a file inside a folder that is protected :



so how can i make the header() methode to open script.jar inside a protected folder.
The folder haves name and pass: blabla,balbla for exempel

thanks for help

Hello, this is my first post I hope you guys can help me out here, basically whats going on is I declare a variable "$diree" then I insert it in a function  but its not getting recognized, here is the code:

$institutea "some directory";
$dire = str_replace(" ","-",$institutea);


   # SETTINGS
   $max_width = 110;
   $max_height = 130;
   
   $per_page = 5;
   $page = $_GET['page'];
   
   $has_previous = false;
   $has_next = false;
   
   function getPictures() {
      global $page, $per_page, $has_previous, $has_next;
      if ( $handle = opendir($diree.'/') ) { // done changes here
         $lightbox = rand();
?><table border="1"><tr><td>
         <form action="pro.php" method="post">
      
            
            
         <?php
         echo "<ul id='pictures'>";
         
         $count = 1;
         $skip = $page * $per_page;

         if ( $skip != 0 )
            $has_previous = true;
         
         while ( $count < $skip && ($file = readdir($handle)) !== false ) {
            if ( !is_dir($file) && ($type = getPictureType($file)) != '' )
               $count++;
         }
         $count = 1;
         while ( $count < $per_page && ($file = readdir($handle)) !== false ) {
            if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) {
            
               if ( ! is_dir($dire .'/') ) {  // done changes here
                  mkdir($dire .'/'); // done changes here
               }
               if ( ! file_exists($dire .'/'.$file) ) {
                  makeThumb( $file, $type );
               }
               
               
               echo '<li>';
               echo '<img src="'.$dire .'/'.$file.'" alt="" / width="110" height="110"><br/><input type="checkbox" name="food[]" value='.$file.'>'; // done changes here
               echo '</li>';
               
               $count++;

            }
         }
         echo '</ul>';
         ?>
            </td></tr><tr><td><div align="center"><input type="submit" value="Add"></div></td></tr>
</tr></table></form>
               <?php
         while ( ($file = readdir($handle)) !== false ) {
            if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) {
               $has_next = true;
               break;
            }
         }
      }
   }
   
   function getPictureType($file) {
      $split = explode($dire .'/', $file);
      $ext = $split[count($split) - 1];
      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 makeThumb( $file, $type ) {
      global $max_width, $max_height;
      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, $dire .'/'.$file);
      } else if ( $type == 'png' ) {
         imagepng($new, $dire .'/'.$file);
      } else if ( $type == 'gif' ) {
         imagegif($new, $dire .'/'.$file);
      }
      imagedestroy($new);
      imagedestroy($src);
   }
?>

I'm trying to copy an existing file using a php script, run on the command line.
I've tried copy() and also a script from the comments on copy() http://www.php.net/manual/en/function.copy.php#102320

No matter how I try to open the file I get " failed to open stream: No such file or directory".
I tried changing the name of the file, and the ext, and both.
The file does exist, and if I run cp old new, the file gets copied fine.

Current code
Code: [Select]
<?php
$source = '/www/documents/myfile.pdf';
$destination = '/www/documents/mypdf.pdf.bak';
copy($source, $destination);
?>

Warning: copy(/www/documents/myfile.pdf): failed to open stream: No such file or directory in /www/cron/historic.php on line 3

Code: [Select]
~$ ls /www/hua/compucom113dev/client/documents/
myfile.pdf
~$ cp /www/documents/myfile.pdf /www/documents/myfile.pdf.bak
~$ ls /www/hua/compucom113dev/client/documents/
myfile.pdf
myfile.pdf.bak

I had chmod in the script at one point to make sure the directory had the permissions needed but it didn't make a difference so I removed it. I can add it back if someone can help explain what to set it to. I was doing 0777 then back to 0755 at the end.

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=320945.0

I have a page with links to images. Currently the images open in a blank browser window, and I think this is too plain.
I don't want to create an HTML file for each image.
I want to have an 'image display php file' - img_temp.php.
Along with the reference to the image file, in the link, I want to also be able to send a <title> to the template file.

I think the code below shows kinda' what I'm trying to do, and also shows that I don't quite have a grasp on how to do it.

Any tips would be appreciated.

Index Page (index.php)
Code: [Select]
<html>
<head>
<title>Gallery Links</title>
</head>

<body>
<!-- This is the page with the links to the images -->
<ul>
  <li><a href="img_temp.php?file=images/01.jpg<?php ?title = "This is image 1"?>">Image 01</a></li>
</ul>
</body>
</html>

Image Display Template (img_temp.php)
Code: [Select]
<html>
<head>
<title><?php echo $title ?></title>
</head>
<body>
<!-- This is the page that displays the sent image and displays the appropriate doc title  -->
 <?php
  //Output the image $file
  echo $file;
?>
</body>
</html>

This is probably an easy one, but I can't figure it out and it's pretty much not searchable.

on a linux-machine i have installed filezilla

the filezilla runs pretty well and all is ok.

now i need to have the passwd that i have stored years ago. The passprhase is stored in a plain in a file called sitemanager.xmlfile


I want to find that file and open it with a terminal command.

find . -name *.sitemanager

well i thought that this will return the file I'm looking for. Now how do I open it automatically, without typing the name?
 

	find . -name *sitemanager.xm | open

This doesn't work. It says it doesn't found the open command.


question: why it does not work on opensuse?

should i use any other command - eg the following:
 

	find . -name *xyz | xargs open

or
 

find . -name *sitemanager.xml | xargs open

or

	find . -name *.xyz -exec open {} \;

and
 

find . -name *.xyz -exec open {} \; .

any and all help will be greatly appreciate

again: what is wanted and needet is to find out the passphrase in the filezilla-configuration

Not sure if this relates more to PHP or Javascript/jQuery, but is there a way to attach a file to an email client (i.e Outlook), when user click on a link/button?

I have a link that generates a PDF and I want it so that, when someone click on a link, it will open up their email client with the PDF already attached, and subject already fill in.

I know, you can use <a href="mailto:..."  but that only opens up the mail client. I also need to attach a file with a subject fill in.

Is this doable with PHP or Javascript?

Hello, I am new to the forum and somewhat new to php, nice to meet you all. This is the first time I have ever really scripted with PHP so I'm still learning about all the tools I have and what I have to call things in PHP.

I have a list of urls and as I loop through each one, I'd like to be able to get information from the webpage. The <title> would be a good start. I also want to know the best way for me to compare data I have.

I'll show the basic code below, but I successfully go through each url in this text file. I put it in a <ul><li> list just fine. So if $url == http://www.youtube.com/file how is the normal way to check and see if the word "youtube" is in $url?

I found preg_match() but I think I'm approaching the whole thing wrong because I get no output. I am an intermediate to somewhat advanced scripter in other languages similar to php, I just need to learn how you do the normal things in PHP.

So I'd like to compare a string "youtube" to a variable '$url'. And I would like to be able to grab the title or other info from the file $url. Here is what I have so far. (Recent research showed me how I should do this with an XML file so I will probably change the .txt to .xml) Can you please tell me what to look for as I have been searching and can't really find a comprehensive answer.

I changed the whole page to an echo trying to fix something last night. Before it was written like..

Code: [Select]
<?php
if ($true) {
$var = value
?>
<html code>The value is <?php $var ?> .</html code>
<?php
}
?>
Code: [Select]
/index.php

<?php
include 'include/header.html';


echo "<div id='wrapper'>
<div id='left'>
<div class='article'>
<br />
<p>";
echo "Today is " . date("l") . ", the " . date("jS") . " of " . date("F") . ".";
$lines = file('data/news.txt');
if ($lines){
foreach ($lines as $line_num => $line) {

$url = htmlspecialchars($line);

//Now I have url. I want to check the url and get the <title> & misc. data.
//if youtube is in $url {html code to embed youtube};
//my attempt was $x = file($url); but I got a lot of 404 and 403 errors.

//now I fill html.
echo "<ul id='menu1' class='auroramenu'>
<li><a href='#'>Story ".$line_num."</a> <a style='display: none;' class='aurorashow' href='#'></a> <a style='display: inline;' class='aurorahide' href='#'></a>
<ul> <br />
<p>".$url."</p><br />
   <li style='text-align:right;'><a href='".$url."' target='_blank'>Read the story.</a>&nbsp;&nbsp;</li> 
</ul> 
</li>
</ul>";
}
}
echo "</p><br />
</div>
</div>
<div id='right'>";
include 'include/sidebar.html';
echo "</div><br class='clr' /></div><br />";
include 'include/footer.html';
?>
Thank you for your help.

Need Suggestion for my codes, I want msg1.php to be open in the frame bottomframe.

if($username=="" || $password=="")
   {      
           echo "<form method=\"post\" action=\"msg1.php\" target=\"bottomframe\" >";
   }

Hey all,

I viewed this tutorial and the guy used $_SERVER['DOCUMENT_ROOT'] to reference a view layout for his specific index file. So I tried to imitate and did exactly what he did, but for me I got the following error:
Code: [Select]
Warning: include(/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php) [function.include]: failed to open stream: No such file or directory in /Users/jmerlino/Sites/diet/index.php on line 22

Warning: include() [function.include]: Failed opening '/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php' for inclusion (include_path='.:/usr/lib/php') in /Users/jmerlino/Sites/diet/index.php on line 22
This is the php code that is causing this error:

include($_SERVER['DOCUMENT_ROOT'].'/'.'diet/views/layouts/'.$controller.'.php');

I'm not sure why it's going to:
/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php
instead of:
/Users/jmerlino/Sites/diet/views/layouts/shop.php

Thanks for any response

I have a PHP file that is executed via batch file very frequently for live updating. This is a sort of "sync" file for reading/writing to a MySQL database. I am able to get it functioning absolutely fine when executed manually via a web browser (and my echo debug lines output the expected information), yet when I run the same PHP file via the command line, it seems to just not be capable of opening any file for reading, so the equivalent variables that are correct when executed on a web browser are blank when echoed through the command line.

Is there a different way of handling reading of files when running a php script via the command line, or should it function exactly the same as when run via a browser?

For instance:

$k = "0";
$line = file("examplefile.log")[$k];
echo $line;

Hi there,

Can anybody help me to write php/javascript code which will allow users to open files directly from web browser into desktop application? Here is the specification:

I have a photo editing business with many people working in Photoshop. I am currently developing a web based application (joblist) using Javascript and PHP which should allow the photoshop designers to browse and open files/images directly from joblist/web browser into photoshop. The reason I want this instead of browsing folder is that I have a database where I store who worked on which file, when and how long it took.

The concept is that, designers will select a file and click on start, as soon as they click on start the original file will open in Photoshop and there will be an entry into database (using PHP). Once they finish the task they will close the file and click on Finish button.

My joblist application will be published in a local server and the file will be open on a local network, so when they save the file it will be saved where the source file is located in (local server). The application should work in both PC and Mac.

I have already done all other part of the application except file opening directly from browser to desktop application functionality.

Anybody can help me to write the code (PHP or Javascript) which can open the file from browser (local server) directly into desktop application e.g. PHotoshop or Illustrator?

Thank you very much I look forward to someone's real help!

Best regards
Mr. Sumon

My images generator comprises an array of image names extracted form an images table from a database using a select statement a random number generator, and a string that builds the correct pathname for the selected file. To select one of the images for display, i generate a random number between 1 and the length of the array. Though the generator is working, I noticed one of the random numbers is throwing up this error: Warning: getimagesize(images/): failed to open stream: No such file or directory in An inspection of the array reveals 2 array elements (representing my number of images) but one array element is NULL ( the first entry in the banner table   image_generator.php

require_once('connection.inc.php');
    $sql = 'SELECT `filename` FROM  banner';
    $result = $mysqli->query($sql, MYSQLI_STORE_RESULT) or die(mysqli_error());
    $row = $result->fetch_array(MYSQLI_ASSOC);//an array of image names
    $count = $result->num_rows;
    for ($i = 1; $i <= $count; ++$i) 
    {
       $row[$i] = $result->fetch_array(MYSQLI_ASSOC);
    }
    $i = rand(1, $count); //a random number generator,
       //The random number is used in the final line to build the correct pathname for the selected file.
       $selectedImage = "images/{$row[$i]['filename']}"; 
      if (file_exists($selectedImage) && is_readable($selectedImage)) 
      {
        $imageSize = getimagesize($selectedImage);
      }

var_dump($row)

array (size=1)
    'filename' => string 'ginsomin2.jpg' (length=13)
    null

RANDON IMAGE DISPLAY

require_once 'image_generator.php';
<div id="banner" class="wrapper clearfix">
      <img src="<?php echo $selectedImage; ?>" alt="banner">
   </div>

Kindly advice how i may proceed from here? Thanks.