最好的方法是使用XMLHttpRequest，这里有更好的例子AJAX - Server Response
像这样创建一个 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；示例：
<div id="myResponse"></div>
修改按钮调用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 $key; ?>', this); return false;">Go</button>
花点时间了解一下 Ajax 调用是如何工作的，如果你不使用表单，那么与服务器通信真的很重要
您可以使用 JQuery，但最好不使用 ;)
编辑
使用表单，您可以这样做：
<form id="formId" action="your_page.php" method="post">
<!-- your table here -->
<input type="hidden" id="key" name="key" value="">
</form>
<script>
var go = function(key) {
document.getElementById('key').value = key;
document.getElementById('formId').submit();
}
</script>
编辑：
将$key 替换为门牌号basename($file, ".mp4")
以及您想要的 Ajax 调用 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
如果密钥存在则进行编码
发送您的新 html 表格。
// 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：
我知道这改编自您的preview question，但这段代码“不正确”，它可以工作，没问题，但可以像这样优化：
来自...
<?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));
?>
...到
<?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));
?>