dongxi0605 2014-12-02 19:10
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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().

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3条回答 默认 最新

  • dongzou1964 2014-12-02 19:18
    关注

    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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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