dongxi0605
2014-12-02 19:10
浏览 310
已采纳

为什么即使在PHP中成功建立连接后，我也无法将图像文件上传到远程FTP服务器？

I want to connect to one remote FTP server and upload image file to that server at some specific location. I'm able to connect to the server but not able to upload the file.

Every time the function ftp_put() is returning false. I debugged the code a lot but I'm not understanding where I'm making the mistake.

Following is my code:

HTML code:

<form method="post" enctype="multipart/form-data" action="xyz.php">
  <input type="file" name="student_image" id="student_image" />                  
</form>

PHP code(xyz.php):

  $t = time();
  $allowed_image_extension = array("jpg","jpeg","gif","png","JPG","JPEG","GIF","PNG");
  if(!empty($_FILES['student_image']['name'])) {
    $ext = pathinfo($_FILES['student_image']['name'], PATHINFO_EXTENSION);     
    $student_image_name = 'student_'.$t.'.'.$ext;
    $_POST['student_name'] = $student_image_name;

    $ftp_server="52.237.5.85"; 
    $ftp_user_name="myservercreds"; 
    $ftp_user_pass="MyServerCreds";

    $file = $_FILES['student_image']['name'];//tobe uploaded 
    $remote_file = "/Students/".$_POST['student_name']; 

    // set up basic connection 
    $conn_id = ftp_connect($ftp_server);  

    // login with username and password 
    $login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); 


    // upload a file 
    if (ftp_put($conn_id, $remote_file, $file, FTP_ASCII)) { 
      echo "successfully uploaded $file
"; 
      exit; 
    } else { 
      echo "There was a problem while uploading $file
"; 
      exit; 
    } 
    // close the connection 
    ftp_close($conn_id);
  }

During debugging I got True and resource everywhere except at one place in a function ftp_put().

图片转代码服务由CSDN问答提供 功能建议

我想连接到一个远程FTP服务器并将图像文件上传到该服务器的某个特定位置。 我能够连接到服务器但无法上传文件。

每次函数 ftp_put（）返回false。 我调试了很多代码，但我不知道我在哪里弄错了。

以下是我的代码：

HTML代码： 

 ＆lt; form method =“post”enctype =“multipart / form-data”action =“xyz.php”＆gt; 
＆lt; input type =“file”name =“student_image”id =“student_image”/＆gt;  
＆lt; / form＆gt; 
   
 
 
  PHP代码（xyz.php）：  
 
 
 <  code> $ t = time（）; 
 $ allowed_image_extension = array（“jpg”，“jpeg”，“gif”，“png”，“JPG”，“JPEG”，“GIF”，“PNG”）; \  n if（！empty（$ _ FILES ['student_image'] ['name']））{
 $ ext = pathinfo（$ _ FILES ['student_image'] ['name']，PATHINFO_EXTENSION）;  
 $ student_image_name ='student _'。$ t。'。'。$ ext; 
 $ _POST ['student_name'] = $ student_image_name; 
 
 $ ftp_server =“52.237.5.85”;  
 $ ftp_user_name =“myservercreds”;  
 $ ftp_user_pass =“MyServerCreds”; 
 
 $ file = $ _FILES ['student_image'] ['name']; //上传
 $ remote_file =“/ Student /".$_POST['student_name'  ]。  
 
 //设置基本连接
 $ conn_id = ftp_connect（$ ftp_server）;  
 
 //使用用户名和密码登录
 $ login_result = ftp_login（$ conn_id，$ ftp_user_name，$ ftp_user_pass）;  
 
 
 //上传文件
 if（ftp_put（$ conn_id，$ remote_file，$ file，FTP_ASCII））{
 echo“已成功上传$ file 
”;  
退出;  
} else {
 echo“上传$ file时出现问题
”;  
退出;  
} 
 //关闭连接
 ftp_close（$ conn_id）; 
} 
   
 
 
在调试过程中，除了在一个地方，我得到了True和资源 在函数 ftp_put（）中。
  • 写回答
  • 好问题 提建议
  • 追加酬金
  • 关注问题
  • 收藏
  • 邀请回答

3条回答 默认 最新

  • dongzou1964 2014-12-02 19:18
    已采纳

    You're using the wrong filename:

    $file = $_FILES['student_image']['name'];//tobe uploaded 
                                ^^^^^^^^

    That is the filename of the file on the USER's computer. PHP doesn't use that. It puts the upload into a random file name, and stores that name in ['tmp_name']. You want

    $file = $_FILES['student_image']['tmp_name'];

    instead. Since you're using the wrong filename, a file which DOESN'T exist on the server, ftp_put is properly returning false since that non-existent file can't be read.

    评论
    解决 无用
    打赏 举报
查看更多回答(2条)

报告相同问题？

提交

相关推荐 更多相似问题