Question

我正在处理使用PHP绑定网站的任务，而PHP的功能是编辑服务器中的HTML文件，然后将其下载给用户。

我尝试过PHP DOM而没有运气。

到目前为止，PHP代码

<a href="download.php?filename=ProjectA.html" class="links">Download ProjectA</a></td>  </tr>  <tr>    <td align="left" valign="middle"><img src="files/tick.jpg" width="16" height="16" /></td>    <td align="left" valign="middle">

我要编辑的HTML代码是以下第137行

<form accept-charset="utf-8" action="Enter your ID"
    method="post">

我的download.php：

<?php

function output_file($file, $name, $mime_type='')
{
 /*
 This function takes a path to a file to output ($file),  the filename that the browser will see ($name) and  the MIME type of the file ($mime_type, optional).
 */

 //Check the file premission
 if(!is_readable($file)) die('File not found or inaccessible!');

 $size = filesize($file);
 $name = rawurldecode($name);

 /* Figure out the MIME type | Check in array */
 $known_mime_types=array(
    "pdf" => "application/pdf",
    "txt" => "text/plain",
    "html" => "text/html",
    "htm" => "text/html",
    "exe" => "application/octet-stream",
    "zip" => "application/zip",
    "doc" => "application/msword",
    "xls" => "application/vnd.ms-excel",
    "ppt" => "application/vnd.ms-powerpoint",
    "gif" => "image/gif",
    "png" => "image/png",
    "jpeg"=> "image/jpg",
    "jpg" =>  "image/jpg",
    "php" => "text/plain"
 );

 if($mime_type==''){
     $file_extension = strtolower(substr(strrchr($file,"."),1));
     if(array_key_exists($file_extension, $known_mime_types)){
        $mime_type=$known_mime_types[$file_extension];
     } else {
        $mime_type="application/force-download";
     };
 };

 //turn off output buffering to decrease cpu usage
 @ob_end_clean(); 

 // required for IE, otherwise Content-Disposition may be ignored
 if(ini_get('zlib.output_compression'))
  ini_set('zlib.output_compression', 'Off');

 header('Content-Type: ' . $mime_type);
 header('Content-Disposition: attachment; filename="'.$name.'"');
 header("Content-Transfer-Encoding: binary");
 header('Accept-Ranges: bytes');

 /* The three lines below basically make the 
    download non-cacheable */
 header("Cache-control: private");
 header('Pragma: private');
 header("Expires: Mon, 26 Jul 1997 05:00:00 GMT");

 // multipart-download and download resuming support
 if(isset($_SERVER['HTTP_RANGE']))
 {
    list($a, $range) = explode("=",$_SERVER['HTTP_RANGE'],2);
    list($range) = explode(",",$range,2);
    list($range, $range_end) = explode("-", $range);
    $range=intval($range);
    if(!$range_end) {
        $range_end=$size-1;
    } else {
        $range_end=intval($range_end);
    }

    $new_length = $range_end-$range+1;
    header("HTTP/1.1 206 Partial Content");
    header("Content-Length: $new_length");
    header("Content-Range: bytes $range-$range_end/$size");
 } else {
    $new_length=$size;
    header("Content-Length: ".$size);
 }

 /* Will output the file itself */
 $chunksize = 1*(1024*1024); //you may want to change this
 $bytes_send = 0;
 if ($file = fopen($file, 'r'))
 {
    if(isset($_SERVER['HTTP_RANGE']))
    fseek($file, $range);

    while(!feof($file) && 
        (!connection_aborted()) && 
        ($bytes_send<$new_length)
          )
    {
        $buffer = fread($file, $chunksize);
        print($buffer); //echo($buffer); // can also possible
        flush();
        $bytes_send += strlen($buffer);
    }
 fclose($file);
 } else
 //If no permissiion
 die('Error - can not open file.');
 //die
die();
}
//Set the time out
set_time_limit(0);

//path to the file
$file_path='files/'.$_REQUEST['filename'];


//Call the download function with file path,file name and file type
output_file($file_path, ''.$_REQUEST['filename'].'', 'text/plain');


?>

我需要协助：（

Answer 1

您需要使用file_get_contents()之类的内容来从文件中获取数据。

以下是伪代码，因为我不知道您的目录/文件结构。

首先，您需要获取文件内容：

<?php 

$file = 'path/to/file';
// check if file exists
if(!file_exists($)) {
    die("File: {$file} does not exist");
} else {
    $data = file_get_contents($file);
}

?>
<html>
<head></head>
<body>
<!-- form stuff here -->
<textarea name="content" cols="10"><?php echo (isset($data))? $data : '';?></textarea>
<!-- more form stuff here -->
</bod>
</html>

现在，当他们提交数据时，您只需使用file_put_contents()。

<?php 

$file = 'path/to/file';

$data = $_POST['content'];
// test if content put successfully
if(!file_put_contents($file, $data)) {
    echo 'failed';
} else {
    echo 'success';
}

?>

PHP以HTML格式保存信息然后下载

1 个答案: