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Sur d’autres sites (10944)
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How to Show result after uploaded file in PHP
23 avril 2022, par i0x4rI have a script that uploads the video to a server, everything is correct but there is a problem, after the upload of the video to the server is completed
it shows all the uploaded files in the (uploads) folder as array !


I only want the result of the file I just uploaded, and it doesn't show me the previous files !
I need ffmpeg to improve video quality


index.php


<?php
//index.php

?>


 
 
 
 <code class="echappe-js"><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
 
<script src="https://cdnjs.cloudflare.com/ajax/libs/dropzone/5.5.1/dropzone.js"></script>
 
 
 



 
 
 








 



<script>&#xA;&#xA;$(document).ready(function(){&#xA; &#xA; Dropzone.options.dropzoneFrom = {&#xA; autoProcessQueue: true,&#xA; timeout: 300000,&#xA; acceptedFiles:"video/*",&#xA; init: function(){&#xA; var submitButton = document.querySelector(&#x27;#submit-all&#x27;);&#xA; myDropzone = this;&#xA; submitButton.addEventListener("click", function(){&#xA; myDropzone.processQueue();&#xA; });&#xA; this.on("complete", function(){&#xA; if(this.getQueuedFiles().length == 0 &amp;&amp; this.getUploadingFiles().length == 0)&#xA; {&#xA; var _this = this;&#xA; _this.removeAllFiles();&#xA; }&#xA; list_image();&#xA; });&#xA; },&#xA; };&#xA;&#xA; list_image();&#xA;&#xA; function list_image()&#xA; {&#xA; $.ajax({&#xA; url:"upload.php",&#xA; success:function(data){&#xA; $("#preview").html(data);&#xA; }&#xA; });&#xA; }&#xA;&#xA; $(document).on(&#x27;click&#x27;, &#x27;.remove_image&#x27;, function(){&#xA; var name = $(this).attr(&#x27;id&#x27;);&#xA; $.ajax({&#xA; url:"upload.php",&#xA; method:"POST",&#xA; data:{name:name},&#xA; success:function(data)&#xA; {&#xA; list_image();&#xA; }&#xA; })&#xA; });&#xA; &#xA;});&#xA;</script>


upload.php


<?php

//upload.php

$folder_name = 'upload/';
$tumb_name = 'thumb/';
$imageext = '.png';

if(!empty($_FILES))
{

 $temp_file = $_FILES['file']['tmp_name'];
 $location = $folder_name . $_FILES['file']['name'];
 move_uploaded_file($temp_file, $location);
 $upload = $_FILES['file']['name'];
 $uploadStr = str_replace(" ", "\ ",$upload);
 $locationStr = str_replace(" ","\ ",$location);
 $cmd = "ffmpeg -y -i {$locationStr} -ss 00:00:15 -vframes 1 thumb/{$uploadStr}.png 2>&1";
 echo shell_exec($cmd);
}

if(isset($_POST["name"]))
{
 $filename = $folder_name.$_POST["name"];
 $imagename = $thumb_name.$_POST["name"].$imageext;
 unlink($filename);
 unlink($imagename);
}

$result = array();

$files = scandir('upload');

$output = '<div class="row">';

if(false !== $files)
{
 foreach($files as $file)
 {
 if('.' != $file && '..' != $file)
 {
 $output .= '
 <a href="http://stackoverflow.com/view.php?file=&#x27;.$file.&#x27;" target="_blank"> <img src="http://stackoverflow.com/feeds/tag/thumb/&#x27;.$file.&#x27;.png" class="img-thumbnail" width='246' height='138' /></a>
 <button type="button" class="btn btn-link remove_image">Remove</button>
 ';
 }
 }
}
$output .= '</div>';
echo $output;

?>


EDIT :
I put the example on an Array, I don't want it, I just want it to show the downloaded video I just uploaded as a result.


EDIT 2 :
There are some who say type $location and it displays the downloaded file, but this does not work !!!
I just tried more than once and with several uses, there is no display where the text is empty


This is an example of that


<?php

//upload.php

$folder_name = 'upload/';
$tumb_name = 'thumb/';
$imageext = '.png';

if(!empty($_FILES))
{

 $temp_file = $_FILES['file']['tmp_name'];
 $location = $folder_name . $_FILES['file']['name'];
 move_uploaded_file($temp_file, $location);
 $upload = $_FILES['file']['name'];
 $uploadStr = str_replace(" ", "\ ",$upload);
 $locationStr = str_replace(" ","\ ",$location);
 $cmd = "ffmpeg -y -i {$locationStr} -ss 00:00:15 -vframes 1 thumb/{$uploadStr}.png 2>&1";
 echo shell_exec($cmd);
}

if(isset($_POST["name"]))
{
 $filename = $folder_name.$_POST["name"];
 $imagename = $thumb_name.$_POST["name"].$imageext;
 unlink($filename);
 unlink($imagename);
}


$output .= 'Successfly file is "'.$location.'"';
echo $output;

?>


Result : Successfly file is ""
no name file :(


EDIT 3 :


this code upload.php
functions not working


<?php

//upload.php

$folder_name = 'upload/';
$tumb_name = 'thumb/';
$imageext = '.png';

if(!empty($_FILES))
{

 $temp_file = $_FILES['file']['tmp_name'];
 $location = $folder_name . $_FILES['file']['name'];
 move_uploaded_file($temp_file, $location);
 $upload = $_FILES['file']['name'];
 $uploadStr = str_replace(" ", "\ ",$upload);
 $locationStr = str_replace(" ","\ ",$location);
 $cmd = "ffmpeg -y -i {$locationStr} -ss 00:00:15 -vframes 1 thumb/{$uploadStr}.png 2>&1";
 echo shell_exec($cmd);
}


echo "The file " . $location . " has been uploaded";
// not working
echo "<br />";
echo "The file " . $upload . " has been uploaded";
// not working
echo "<br />";
echo "The file " . $uploadStr . " has been uploaded";
// not working
echo "<br />";
echo "The file " . $locationStr . " has been uploaded";
// not working

?>




this error :
[23-Apr-2022 12:31:56 Asia/Riyadh] PHP Notice : Undefined variable : location in /home/prdix/public_html/test/upload.php on line 38


line 38 : echo $location ;


About the developer solution Markus AO


I did the experiment and it is quite good, but dropzone is still missing, because I will upload a large video and the normal upload compresses the video before uploading, here is a picture from my mobile while uploading, but this does not happen with dropzone



I want to implement this in dropzone as well, because this library does not compress the video, but upload it in full size.


Thank you bro.


-
FFmpeg slide show duration from text file
8 mai 2022, par krvI am using ffmpeg to create a slideshow of images from a list of files. I am using the following command.


ffmpeg -f concat -i List.txt output.mp4


How can is set the duration of all the images in the list to 5 seconds so each image is displayed for 5 seconds ?


Thanks


-
Can't show image with opencv when importing av
5 mars, par FlojomojoWhen importing the PyAv module, I am unable to show an image with opencv using imshow()


Code without the PyAv module (works as expected)


import cv2

img = cv2.imread("test_image.jpeg")
cv2.imshow('image', img)
cv2.waitKey(0)


Code with the import (doesn't work, just hangs)


import cv2
import av

img = cv2.imread("test_image.jpeg")
cv2.imshow('image', img)
cv2.waitKey(0)


OS : Linux arch 5.18.3-arch1-1 #1 SMP PREEMPT_DYNAMIC Thu, 09 Jun 2022 16:14:10 +0000 x86_64 GNU/Linux


Am I doing something wrong or is this a (un-)known issue ?


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