php - 如何使用 ffmpeg/php 在按钮单击时将 mp4 文件转换为 mp3？

Question

我正在编写如下所示的 php 代码，我正在转换 mp4 文件转为 mp3 使用系统命令 ffmpeg (在下面的案例陈述中) .

<?php 

$mp4_files = preg_grep('~\.(mp4)$~', scandir($src_dir)); 

foreach ($mp4_files as $f)
 {

     $parts = pathinfo($f);
     switch ($parts['extension'])
     {
         case 'mp4' :
             $filePath = $src_dir . DS . $f;
             system('ffmpeg -i ' . $filePath . ' -map 0:2 -ac 1 ' . $destination_dir . DS . $parts['filename'] . '.mp3', $result);  // Through this command conversion happens. 
     }
 }

$mp3_files = preg_grep('/^([^.])/', scandir($destination_dir));

?>

转换后， mp3 文件进入destination_dir .如果新的 mp4 文件到达 $src_dir ，转换通常发生在页面刷新时。

转换完成后，我将所有内容解析为表格，如下所示:

<table>
   <tr>
      <th style="width:8%; text-align:center;">House Number</th>
      <th style="width:8%; text-align:center;">MP4 Name</th>
      <th style="width:8%; text-align:center;" >Action/Status</th>
   </tr>
   <?php
      $mp4_files = array_values($mp4_files);
      $mp3_files = array_values($mp3_files);
      foreach ($programs as $key => $program)    { 
         $file = $mp4_files[$key];     
         $file2 = $mp3_files[$key];   // file2 is in mp3 folder
      ?>
   <tr>
      <td style="width:5%; text-align:center;"><span style="border: 1px solid black; padding:5px;"><?php echo basename($file, ".mp4"); ?></span></td> <!-- House Number -->
      <td style="width:5%; text-align:center;"><span style="border: 1px solid black; padding:5px;"><?php echo basename($file); ?></span></td> <!-- MP4 Name -->             
      <td style="width:5%; text-align:center;"><button style="width:90px;" type="button" class="btn btn-outline-primary">Go</button</td>  <!-- Go Button -->
   </tr>
   <?php } ?>
</table>

问题陈述:

我想知道单击 时应该在上面的 php 代码中进行哪些更改前往按钮 , 个人转换 mp4转mp3 发生。

点击 前往按钮 , 属于单个行的单个 mp3 文件(来自 mp4)应该进去 目标目录 ($destination_dir) .

Answer 1

最好的方法是使用 XMLHttpRequest，这里有更好的例子 AJAX - Server Response

创建一个像这样的javascript函数:

<script>
  // Check if the window is loaded
  window.addEventListener('load', function () {

    // Function to call Ajax request to convert or move file
    var go = function(key, btn) {

      // Initialize request
      var xhttp = new XMLHttpRequest();

      // Execute code when the request ready state is changed and handle response.
      // Optional but recommended.
      xhttp.onreadystatechange = function() {
        if (this.readyState == 4 && this.status == 200) {
          // Do what you want here with the response here
          document.getElementById('myResponse').innerHTML = this.responseText;

          // Disable the button to not clicking again
          // see https://www.w3schools.com/jsref/prop_pushbutton_disabled.asp
          btn.disabled = true;
         }
      };

      // Handle error message here
      // Optional but recommended.
      xhttp.onerror = function(event) {
        document.getElementById('myResponse').innerHTML = 'Request error:' + event.target.status;
      };

      // Create request to the server
      // Call the page that convert .mp4 or move .mp3
      xhttp.open('POST', '/your_convert_file.php', true);

      // Pass key or name or something (secure) to retrieve the file
      // and send the request to the server
      xhttp.send('key=' + key);
    }
 )};
</script>

根据需要添加一些东西来处理服务器的响应；例子:

<div id="myResponse"></div>

修改按钮以调用 javascript 函数 onclick="go('<?php echo $key; ?>', this); return false;" :

<button style="width:90px;" type="button" class="btn btn-outline-primary" onclick="go('<?php echo $key; ?>', this); return false;">Go</button>

花点时间了解 Ajax 调用的工作原理，如果您不使用表单，与服务器通信非常重要

您可以使用 JQuery，但最好不使用 ;)

编辑

使用表单，您可以这样做:

<form id="formId" action="your_page.php" method="post">

<!-- your table here -->

<input type="hidden" id="key" name="key" value="">
</form>

<script>
  var go = function(key) {
    document.getElementById('key').value = key;
    document.getElementById('formId').submit();
  }
</script>

编辑 :

用门牌号 $key 替换 basename($file, ".mp4")
以及 Ajax 调用所需的 page.php 或 your_encoder.php :

// EXAMPLE FOR AJAX CALL

<?php 
// Get the unique name or key
$key = $_POST['key'];

// If key is empty, no need to go further.
if(empty($_POST['key'])) {
  echo "File name is empty !";
  exit();
}

// Can be secure by performing string sanitize
$filePath = $src_dir . DS . $key . '.mp4';

// Check if file exists
// echo a json string to parse it in javascript is better
if (file_exists($filePath)) {
    system('ffmpeg -i ' . $filePath . ' -map 0:2 -ac 1 ' . $destination_dir . DS . $parts['filename'] . '.mp3', $result);
    echo "The file $filePath has been encoded successfully.";
      . "<br />"
      . $result;
} else {
    echo "The file $filePath does not exist";
}
?>

如果你使用 form ，你必须:

检查$_POST['key']是否存在

如果 key 存在则进行编码

发送您的新 html 表。

// EXAMPLE FOR FORM CALL

<?php
// Get the unique name or key
$key = $_POST['key'];

// If key is not empty.
if(!empty($_POST['key'])) {
  // do the encoding here like above
  // set message success | error
}

// display your html table and message here.
?>

编辑 :

我知道这是根据您的 preview question 改编的，但是此代码“不正确”，可以正常工作，没问题，但是可以像这样进行优化:

从...

<?php 
// Here, you list only .mp4 in the directory
// see: https://www.php.net/manual/en/function.preg-grep.php
$mp4_files = preg_grep('~\.(mp4)$~', scandir($src_dir)); 

// Here you loop only on all .mp4 
foreach ($mp4_files as $f)
 {
     $parts = pathinfo($f);

     // Here, you check if extension is .mp4
     // Useless, because it is always the case.
     // see : https://www.php.net/manual/en/control-structures.switch.php
     switch ($parts['extension'])
     {
         case 'mp4' :
             $filePath = $src_dir . DS . $f;
             system('ffmpeg -i ' . $filePath . ' -map 0:2 -ac 1 ' . $destination_dir . DS . $parts['filename'] . '.mp3', $result);  // Through this command conversion happens. 
     }
 }

$mp3_files = preg_grep('/^([^.])/', scandir($destination_dir));
?>

... 至

<?php
// Here, you list only .mp4 on the directory
$mp4_files = preg_grep('~\.(mp4)$~', scandir($src_dir)); 

// Here you loop only on all .mp4 
foreach ($mp4_files as $f)
 {
     $filePath = $src_dir . DS . $f;

     // No more need to switch, preg_reg do the job before looping
     // Through this command conversion happens.
     system('ffmpeg -i ' . $filePath . ' -map 0:2 -ac 1 ' . $destination_dir . DS . pathinfo($f, 'filename') . '.mp3', $result);  
 }

$mp3_files = preg_grep('/^([^.])/', scandir($destination_dir));
?>