PHP - Warning: Readfile() [function.readfile]: Help Please

I wrote a small application to force downloads of various large video files using readfile. This worked fine. However, the users have changed hosts and now it doesn't work. Small files are fine, but large files will cut off before finishing. Test file is 114MB, but it is always cut off at exactly 67.2MB or 51.4MB depending on a couple criteria.

I've stripped the application down to the key part that's not working and run it on a test page the just immediately goes the file - no logins or any of that shenanigans and htaccess blocking either:

$file = some/file/on/the/server.wmv; //114MB
hea der('Content-Description: File Transfer');
hea der("Content-Type: application/octet-stream");
hea der("Content-Disposition: attachment; filename=" .basename($file));
hea der("Content-Transfer-Encoding: binary");
hea der('Expires: 0');
hea der('Cache-Control: must-revalidate, post-check=0, pre-check=0');
hea der('Pragma: public');
hea der('Accept-Ranges: bytes');
hea der("Content-Length: ".filesize($file));
ob_clean();
flush();
@readfile($file);

(gaps in the header words are to prevent forum breakage)

I've tried the following variations:
transfer encoding - chunked (this results in getting a 51.4 MB download rather than 67.2) content-type application/force-download tried the default force-download code from the php.net manual tried the chunking function variations on php.net readfile page tried without accept ranges originally (only spotted that on this site) set_time_limt to one hour set max memory to 300MB (phpinfo has shown these changes did get accepted) used htacess to disable gzip and deflate (this results in getting a 51.4 MB download rather than 67.2)
(SetEnv no-gzip dont-vary
or
RewriteEngine On
RewriteRule . - [E=no-gzip:1]
or
RemoveOutputFilter DEFLATE html txt xml css js php wmv)
None of these have solved the issue. The code worked fine on the old site, works fine on my site, and works fine on my test server. Direct downloads from the broken site also work fine, just not readfile downloads. The filesize is reading correctly if I echo just that, and it reads properly on the progress bar when downloading through readfile.

The hosting company are blaming my coding. The htaccess is the bit I'm least sure of and I find it hard to get solid info on this part on the web.

Is there anything wrong with my code that could be causing this? Am I missing something? What could be causing the problem?

Many Thanks

In a PHP file [php_one.php], I wrote some PHP code that contained a form.  I used ECHO to write the HTML inside of the PHP tags.

On another PHP page [php_two.php], I wanted to show the code that was used to create the first page [php_one.php].

My first effort provided an interesting result, since the PHP code was readable.  However, all HTML code was visible in its executed form.

There was an actual input field and a submit button (not the code used to produce them) in the context of my coding.

I have tried using:

file_get_contents("FILE")
$array = file("FILE")
readfile("FILE")

All results were similar and carried the same problem.

How can I view code so that even the HTML is visible as code?

An image is saved in the root directory.
The page "label.php" uses "readfile" to access and display the image using code like this:

$file = '../image.jpg'; // IN ROOT DIRECTORY
$type = 'image/jpeg';
header('Content-Type:'.$type);
header('Content-Length: ' . filesize($file));
readfile($file);

The image then gets embedded into an html page and displayed like this:
Dear Customer, here is your image: <img src="label.php?label=<?= $filename ?>.jpg">

QUESTION: in the event of an error, how do I dipslay the readfile error on the customer's html page?
Even though I have error trapping in the "readfile" code in "label.php" file -- like if a parameter is missing, I'll have exit( "NO IMAGE GENERATED" ).  But this error message won't display on the customer's html page because html treats <?= $filename ?>.jpg" like it's a real image instead of displaying the readfile's "NO IMAGE GENERATED" exit error.

Make sense? No. I probably have to rewrite my question 😃

Edited Monday at 01:07 AM by ChenXiu

I have an FLV player and I am using a php file to read an FLV that is above webroot. It reads the FLV just fine, however, during its cache process or while it is still downloading, if I click on the home button (home button of the current site), the server does not respond until the FLV has been fully cached.

Now this is only occuring in firefox, chrome and IE. In safari it works fine. Who know, maybe it is a header issue. Here is my script that I pass to my FLV player

<?php
include("connect.php");
if ($_SESSION['loggedIn'] == true)
{
    $id        = $_GET['id'];

    $query    = mysql_query("SELECT video FROM reviews WHERE id = '$id'");
    $row    = mysql_fetch_array($query);

    $filename    = "../../../armsmedia/videos/".$row['video'];

    header( 'Content-Description: File Transfer' );
    header( 'Content-Type: application/octet-stream' );
    header( 'Content-Disposition: attachment; filename='.basename( $filename ) );
    header( 'Content-Transfer-Encoding: binary' );
    header( 'Expires: 0' );
    header( 'Cache-Control: must-revalidate, post-check=0, pre-check=0' );
    header( 'Pragma: public' );
    header( 'Content-Length: ' . filesize( $filename ) );
    ob_clean();
    flush();
    readfile( $filename );
    exit;
}
?>

An image is secured in the root directory, to be displayed to customer in html document.

This image can be displayed as a link to a php readfile page
The readfile page uses getimagesize("../rootdir/abc.png"), headers, and readfile()
echo "<img src='READFILE.php?image_name=abcd.png'>";
-or-
The image can be displayed directly to browser using
file_get_contents("../rootdir/abc.png"), base64_encode(), and data:image;base64.
echo '<img src="data:image;base64, ' .$img. ' " style="width:5in;height:5in;">';

Which is better? Why and why not?

Thank you.

Edited yesterday at 01:15 PM by ChenXiu

Which is better? Why and why not?

An image is secured in the root directory, to be displayed to customer in html document.

This image can be displayed as a link to a php readfile page
The readfile page uses getimagesize("../rootdir/abc.png"), headers, and readfile()
echo "<img src='READFILE.php?image_name=abcd.png'>";
-or-
The image can be displayed directly to browser using
file_get_contents("../rootdir/abc.png"), base64_encode(), and data:image;base64.
echo '<img src="data:image;base64, ' .$img. ' " style="width:5in;height:5in;">';

Thank you.

I am wanting HTML to render within a table and sourcing the data from another file. In an attempt to do this, I tried:
Code: [Select]
<TABLE class="main">
                <TR>
                        <TD>
                                <?php
                                        readfile('./1/index.html');
                                ?>
                        </TD>
                </TR>
        </TABLE>
This does not work and I am wondering if anyone knows how it can be done - or even if it can be done?

Circumstance:
"label.php" is actually PHP "readfile" page that fetches images from a private root directory.
"customer.html" is the public page where visitors view their images like this:
Dear Customer, Here Is Your Image: <img src="label.php?image=id_number_1001.png">

Question:
Is it okay to use a hidden $_SESSION["filename"] and change what I have now:
Dear Customer, Here Is Your Image: <img src="label.php?image=id_number_1001.png">
To this:
Dear Customer, Here Is Your Image: <img src="label.php">

This works by having the actual image name stored in a hidden session variable, using the following lines of code on my readfile page "label.php" like this:
$imgName = $_SESSION[ "filename" ] . ".png";
$imgPath = "../root_directory/" . $imgName;

Since I do not want customers to be allowed to ever see the image again after logging out, is it okay to rely on a hidden session variable for the filename as I described above?

(Potential hackers will be less tempted to "try different image names and numbers" like "id1.png, id2.png, id3.png" because they will have no idea if the image will be a gif, jpeg, or png, nor have any idea of a numbering sequence.)

Thank you!!

 

Have a look and please explain

Im making a site for my family to leave messages for everyone to see but every now and then the message wont load and i get theses errors,

Warning: mysql_connect() [function.mysql-connect]: Too many connections in /home/#####/public_html/#####.com/messages.php on line 17

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 18

Warning: mysql_query() [function.mysql-query]: Access denied for user '#####'@'localhost' (using password: NO) in /home/#####/public_html/#####.com/messages.php on line 21

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /home/#####/public_html/#####.com/messages.php on line 21

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 24

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/#####/public_html/#####.com/messages.php on line 25

Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in /home/#####/public_html/#####.com/messages.php on line 54


i also recieved this error once

Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock' (2)

The hosting support team have been no help at all.

Hello

I'm using tcpdf to convert some stuff to pdf.
On my local computer running wamp my script works perfect and the pdf is generated without problems
When i put my files in a webserver i got this error:
Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/vieira/public_html/topdf/dbcon.class.php on line 68

Line 68 is:
if(in_array($key,$arr_switches)){   //check switches

My complete code is:

    // Get fields of enum on/off switches type
    $result2 = mysql_query("SHOW FIELDS FROM $db_name.$table");
    $counter=0;    
    while($row = @mysql_fetch_array($result2)){
        //echo $row['Field'] . ' ' . $row['Type']."<br/>";
        if($row['Type'] === "enum('on','off')"){
            $arr_switches[$counter] = $row['Field'];
            $counter++;
        }        
    }
    //print"<pre>"; print_r($arr_switches); print"</pre>";exit;
    
    $counter = 0;
    while($row = @mysql_fetch_array($result)){
        foreach( $row as $key => $val ){
            if(!is_numeric($key))
            {
                $row_rs_certidao[$key] = $val;
                if(in_array($key,$arr_switches)){   //check switches
                    $record_key[$counter] = htmlentities('<?php if (!(strcmp($row_rs_certidao['."'".$key."'".'],"on"))) {echo "x";} ?>');
                    if($val==='on') $record_val[$counter] = "x";    //turn on switches
                    else $record_val[$counter] = '';                //turn off
                }else{
                    $record_key[$counter] = htmlentities('<?php echo $row_rs_certidao['."'".$key."'".']; ?>');
                    $record_val[$counter] = htmlentities($val);
                }
                $counter++;                
            }            
        }        
    }

Anyone can help?

Please help me to solve this error

Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) in C:\wamp\www\shop-script-free\includes\database\mysql.php on line 13

Warning: mysql_get_server_info() expects parameter 1 to be resource, boolean given in C:\wamp\www\shop-script-free\includes\database\mysql.php on line 14
Access denied for user 'SYSTEM'@'localhost' (using password: NO)

Thanks you very much

Please help me with this code. It keep appear in my website.

"Warning: preg_match() [function.preg-match]: No ending delimiter '.' found in /home/kengsite/public_html/wp-content/plugins/statpress/statpress.php on line 1184"

And this is how the line look like.

if (!preg_match(".ico$", $urlRequested)) { return ''; }

Hi All,

I am trying to create a PHP code that will run a form, however i keep getting the error message

Warning: preg_match() [function.preg-match]: No ending delimiter '^' found in D:\xampp\htdocs\HFCC\Scripts\mailprocess.php on line 55

Invalid email address entered. Go Back and try again

attached is the complete copy of the code i am using, i have an email text box and label with the id 'email' also my method is POST and my action is Scripts/mailprocess.php.

Please let me know if you need any additional details. Any help will be much appreciated.

Thanks