I have the following code which kind of works but for some reason the $download includes the part inside into the downloaded file.
<form method="post">
<button id="click" name="click">Download</button>
</form>
<?php
if(isset($_POST['click'])){
$files = scandir('/local/path', SCANDIR_SORT_DESCENDING);
$newest_file = $files[0];
$download = $newest_file;
header("Content-Type: text/plain");
header('Content-Disposition: attachment; filename="'.$download.'"');
readfile("/local/path/$download");
}
?>
Downloaded file
form method="post">
<button id="click" name="click">Download</button>
</form>
Start-of-the-actual-log
blabla
blabla
blabla
What is causing this to happen? I only want to download the newest file in a folder.
Dont print anything or send header before downloading, look at this :
<?php
if(isset($_POST['click'])){
$files = scandir('/local/path', SCANDIR_SORT_DESCENDING);
$newest_file = $files[0];
$download = $newest_file;
header("Content-Type: text/plain");
header('Content-Disposition: attachment; filename="'.$download.'"');
readfile("/local/path/$download");
}else{
echo'
<form method="post">
<button id="click" name="click">Download</button>
</form>';
}
?>
Related
Why when i try to export MySql table to CSV with header('Content-Disposition:attachment; filename="'.$filename.'"'); It doesn't get done properly:
it does create the CSV file
however it does it on the beginning of the file, where the page code is
and after the code the is table content
This is the code witch exports it:
$this->view->table = $model->info('name');
$is_csv = $this->_getParam('csv');
if ($this->_request->isPost() && $is_csv) {
$fichier = 'file.csv';
header('Content-type: text/plain');
header('Content-Disposition: attachment; filename="'.$fichier.'"');
$fp = fopen('php://output', 'w+');
$data = $model->fetchAll();
foreach ($data as $fields) {
fputcsv($fp, $fields->toArray());
}
fclose($fp);
}
And here i'm calling this with button:
<form method="post">
<input type="hidden" name="table" value="<?php echo $this->table ?>" />
<button type="submit" class="btn" name="csv" value="csv">
<?php echo Core_Locale::translate('CSV')?>
</button>
</form>
put this code up in your application then application create view.
I had same issue but i solved it this way..
i put peace of code to function/block and place it over echo or render html
put exit; on the end of function/block
if(isset($_POST['submit'])) {
$csv = new Csv();
$filename = $csv->generateFileDate();
if($filename !== false) {
$data = file_get_contents($csv->folder .'reports/'. $filename);
header('Content-Type: application/csv; charset=utf-8');
header('Content-Disposition: attachement;filename="'.$filename.'";');
echo $data;
exit;
}
}
?>
<!doctype html>
<html lang="us">
<head> .........
hope this help to you
I am making a downloader for mp3 and mp4 and i want two php files in php with 2 buttons calling both php files but only the first php file works.
code for popup.php
<?php
include "downloadmp3.php";
include "downloadmp4.php";
?>
<html>
<head></head>
<body bgcolor="#E6E6E6">
<form method="post">
<label for="url">Download mp3:</label>
<input type="text" name="url" value="" id="url">
<input type="submit" name="submit" value="Download">
<hr>
</form>
<form method="post">
<label for="url1">Download mp4:</label>
<input type="text" name="url1" value="" id="url1">
<input type="submit" name="submit" value="Download">
</form>
</body>
</html>
code for downloadmp3.php
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$url = (isset($_POST['url']) && !empty($_POST['url'])) ? $_POST['url'] : false;
if (!$url) {
echo "Vul alstublieft een url in";
} else {
$source = file_get_contents($url);
$source = urldecode($source);
// Verkrijg de video titel.
$vTitle_results_1 = explode('<title>', $source);
$vTitle_results_2 = explode('</title>', $vTitle_results_1[1]);
$title = trim(str_replace(' – YouTube', ”, trim($vTitle_results_2[0])));
// Extract video download URL.
$dURL_results_1 = explode('url_encoded_fmt_stream_map', "url=", $source);
$dURL_results_2 = explode('\u0026quality', $dURL_results_1[1]);
// Force download van d video.
$file = str_replace(' ', '_', strtolower($title)).'.mp4';
header("Cache-Control: public");
header("Content-Description: File Transfer");
header("Content-Disposition: attachment; filename=$file");
header("Content-Type: video/mp4");
header("Content-Transfer-Encoding: binary");
readfile($dURL_results_2[0]);
exit;
}
}
?>
and code for downloadmp4.php
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
$url = (isset($_POST['url1']) && !empty($_POST['url1'])) ? $_POST['url1'] : false;
if (!$url) {
echo "Vul alstublieft een url in";
} else {
$source = file_get_contents($url);
$source = urldecode($source);
// Verkrijg de video titel.
$vTitle_results_1 = explode('<title>', $source);
$vTitle_results_2 = explode('</title>', $vTitle_results_1[1]);
$title = trim(str_replace(' – YouTube', ”, trim($vTitle_results_2[0])));
// Extract video download URL.
$dURL_results_1 = explode('url_encoded_fmt_stream_map', "url1=", $source);
$dURL_results_2 = explode('\u0026quality', $dURL_results_1[1]);
// Force download van d video.
$file = str_replace(' ', '_', strtolower($title)).'.mp4';
header("Cache-Control: public");
header("Content-Description: File Transfer");
header("Content-Disposition: attachment; filename=$file");
header("Content-Type: video/mp4");
header("Content-Transfer-Encoding: binary");
readfile($dURL_results_2[0]);
exit;
}
}
?>
Updated:
Form submitted for MP3 form works.
But form submitted for MP4 doesn't works.
Please provide an solution to make it work.
Only first php file calls are working because of the following if condition:
if ($_SERVER['REQUEST_METHOD'] == 'POST'){
Since downloadmp3.php has been included first therefore it's if condition is giving the result.
Use proper form handler to make it work.
Also the best practice is to use unique names for the forms.
This code is working fine for me in all the browsers and the downloaded file is playing good :
download-audio.php
<?php
$file = $_GET['file'];
header ('Content-type: octet/stream');
header ('Content-disposition: attachment; filename='.$file.';');
header('Content-Length: '.filesize($file));
readfile($file);
exit;
?>
test.html
<a href='download-audio.php?file=duetsong.mp3'>Download Duet song MP3</a>
<a href='download-audio.php?file=duetsong.mp4'>Download Duet song MP4</a>
I am working on a project where the requirement is to download a textarea content as a .txt file on a wordpress site.
After searching previous questions, I got below code which marked as a correct answer.
<html>
<body>
<form action="#" method="post">
<textarea name="text" rows="20" cols="100"></textarea>
<input type="submit" value="submit">Download Text</input>
</form>
<?PHP
if(isset($_POST['submit']))
{
$text = $_POST['text'];
print ($text);
$filename = 'test.txt';
$string = $text;
$fp = fopen($filename, "w");
fwrite($fp, $string);
fclose($fp);
header('Content-disposition: attachment; filename=test.txt');
header('Content-type: application/txt');
readfile('test.txt');
}
?>
</body>
</html>
But for me it is not creating a .txt file. please help me to identify the issue
Modified little bit in your code.
First put your php code
<?php
if(isset($_POST['submit']))
{
$text = $_POST['text'];
print ($text);
$filename = 'test.txt';
$string = $text;
$fp = fopen($filename, "w");
fwrite($fp, $string);
fclose($fp);
header('Content-disposition: attachment; filename=test.txt');
header('Content-type: application/txt');
readfile('test.txt');
die; //modified code
}
?>
after that put your html code
<html>
<body>
<form action="#" method="post">
<textarea name="text" rows="20" cols="100"></textarea>
<input type="submit" value="submit" name="submit">Download Text</input>
</form>
add name attribute in submit button.
Your code actually creates a file on the server and it does not get deleted afterwards. If the code failed, it is possible that you didn't have permission to write on your server. You can trim your code down to the following so as no file is generated in the process:
<?php
if(isset($_POST['text']))
{
header('Content-disposition: attachment; filename=test.txt');
header('Content-type: application/txt');
echo $_POST['text'];
exit; //stop writing
}
?>
<html>
<body>
<form action="" method="post">
<textarea name="text" rows="20" cols="100"></textarea>
<input type="submit" value="submit" name="submit">Download Text</input>
</form>
</body>
</html>
Firstly, you should output the file before printing anything on screen and secondly, you forgot to add name="submit" to your original code. I didn't use it here, but I included for you to get an idea.
In the WordPress Context:
Add this to your functions.php or plugin file:
function feed_text_download(){
if(isset($_POST['text_to_download']))
{
header('Content-disposition: attachment; filename=test.txt');
header('Content-type: application/txt');
echo $_POST['text_to_download'];
exit; //stop writing
}
}
add_action('after_setup_theme', 'feed_text_download');
Add the form to one of your template files or in the HTML editor of your post:
<form action="" method="post">
<textarea name="text_to_download" rows="20" cols="100"></textarea>
<input type="submit" value="submit" name="submit">Download Text</input>
</form>
It should be good :)
EDIT: Add Filename
HTML:
<form action="" method="post">
<label>Filename:<input type="text" name="filename" /></label>
<label>Text:<textarea cols="100" name="text_to_download" rows="20"></textarea></label>
<input type="submit" name="submit" value="Download Text" />
</form>
PHP:
function feed_text_download() {
if ( isset( $_POST['text_to_download'] ) && isset( $_POST['filename'] ) ) {
$filename = sanitize_file_name( $_POST['filename'] );
header( 'Content-disposition: attachment; filename=' . $filename );
header( 'Content-type: application/txt' );
echo $_POST['text_to_download'];
exit; //stop writing
}
}
add_action( 'after_setup_theme', 'feed_text_download' );
Try following code, that will help you:
<?php ob_start();?>
<html>
<body>
<?php
if(isset($_POST['tarea'])){
$filename = 'test.txt';
$data = $_POST['tarea'];
header("Cache-Control: public");
header("Content-Description: File Transfer");
header("Content-Disposition: attachment; filename=$filename");
header("Content-Type: application/octet-stream; ");
header("Content-Transfer-Encoding: binary");
echo $data;
exit;
}
?>
<form action="" method="post">
<textarea name="tarea"></textarea>
<input type="submit">
</form>
</body>
</html>
In my website I want the users to download a music file when a button is clicked. The problem I am experiencing is that when the button is clicked the download starts but only 506 KB on a total of 4 MB are downloaded and when it is opened it shows unsupported format. I use this php download script with html form to make the download:
HTML FORM
<FORM ACTION="download.php" METHOD=POST>
<INPUT TYPE="hidden" NAME="music" VALUE= "music/<? print $file; ?>" >
<INPUT TYPE="SUBMIT" NAME="download" VALUE="Download">
</FORM>
PHP DOWNLOAD SCRIPT
<?php
$FileName = $_POST["music"];
$FilePath = "music";
$size = filesize($FilePath . $FileName);
header("Content-Type: application/force-download; name=\"". $FileName ."\"");
header("Content-Transfer-Encoding: binary");
header("Content-Length: ". $size ."");
header("Content-Disposition: attachment; filename=\"". $FileName ."\"");
header("Expires: 0");
header("Cache-Control: private");
header("Pragma: private");
echo (readfile($FilePath . $FileName));
?>
Can you please point me to the problem?
Your code is looking for a file in a location like:
musicmusic/file.mp3
Assuming your file is actually stored in something like music/file.mp3, and variable $file contains something like file.mp3, you want to replace:
<INPUT TYPE="hidden" NAME="music" VALUE= "music/<? print $file; ?>">
with:
<INPUT TYPE="hidden" NAME="music" VALUE= "<? print $file; ?>">
and then replace this:
$FilePath = "music";
with:
$FilePath = "music/";
(or delete $FilePath at all)
Also, I would recommend you to check $_POST["music"] doesn't contain .. or start with / before you let anyone download any file from your server.
I've created a PHP file which shows the results from a MySQL database as we all create like in
echo "<table><tr><td>";
...
echo "</td></tr></table>";
But now I want to make a button at the bottom of the table, something like 'save report', which saves the created table into HTML format.
So how could it be done?
You can use the following scripts:
index.php
In index.php, you have the HTML table.
<?php
$contents = "<table><tr><td>A</td><td>B</td></tr><tr><td>One</td><td>Two</td></tr><tr><td>Three</td><td>Four</td></tr></table>"; // Put here the source code of your table.
?>
<html>
<head>
<title>Save the file!</title>
</head>
<body>
<?php echo $contents; ?>
<form action="savefile.php" method="post">
<input type="hidden" name="contents" value="<?php echo htmlspecialchars($contents); ?>">
<input type="submit" value="Save file" />
</form>
</body>
</html>
savefile.php
And then use the file savefile.php to popup your browser's download dialog to save the file.
<?php
if ($_SERVER['REQUEST_METHOD'] == "POST") {
header('Content-type: text/html');
header('Content-Disposition: attachment; filename="table.html"');
echo $_POST['contents'];
}
?>
I think you want to save reports generated by PHP/MySQL as HTML file?
<?php
// Open file for writing
$fileHandle = fopen("/DestinationPath/DestinationFile.html", "w");
// Dump File
$head = "<html><head><title>my reports</title></head><body>\n";
fwrite($fileHandle, $head);
$sql = mysql_query("Your sql query here");
while ($result = mysql_fetch_assoc($sql)) {
$line = $result['yourmysqlfield']."\n";
fwrite($fileHandle, $line);
}
$foot = "</body></html>\n";
fwrite($fileHandle, $foot);
// Close File
close($fileHandle);
?>