got stuck - uploading via Curl an data-binary option












1















first i've to say, that i'm totally new to cURL. But I've just code some lines with curl and PHP to Upload some XML files.
The following Code is working very good. (just looked around and build up a simple working solution)



Now i've got the issue to use the curl --Data-Binary option instead of the -F (Form) option.



The following Code is in the php-File (curlupload.php):



<?php

@$authcode = htmlspecialchars($_POST["acode"]);

if ($authcode == "1234"){
echo "AuthCode okn";

$upstatus = "1";
$uploadpath = "files/";
$filedata = $_FILES['file']['tmp_name'];
$target_file = $uploadpath . basename($_FILES["file"]["name"]);

#echo "Acode:" . $authcode . "n";
#echo "filedata= ".$filedata."n";
echo "target_file= ".$target_file."n";
#echo "1 upstatus= ".$upstatus."n";

if (strpos($target_file, ".xml") == true) {
$upstatus = "1";
echo "2 upstatus= ".$upstatus."n";

if (file_exists($target_file)) {
echo "Sorry, file already exists.n";
$upstatus = "0";
}

#echo "4 upstatus= ".$upstatus."n";
if ($upstatus == "1") {
if ($filedata != '')
copy($filedata,$target_file);
echo "n-----n";
echo "successn";
}

}else{
$upstatus = "0";
#echo "3 upstatus= ".$upstatus."n";
echo "Filenamecheck Failed.n";
}



if ($upstatus == "0") {
echo "Sorry, your file was not uploaded.n";
}
}else{
echo "NO.";
}
?>


my cURL command for uploading a file is the following:



curl -v -F 'file=@/root/testfile01.xml' -F 'acode=1234' http://webserver/upload2/curlupload.php


Now i'm going to use the -d and --data-binary option like this:



curl --data-binary "file=@testfile02.xml" --data "acode=1234" http://webserver/upload2/curluploadtest.php


with the 2nd above command i receive the following messages:



AuthCode ok
target_file= files/
Filenamecheck Failed.
Sorry, your file was not uploaded.


so, the variables got no input - Can somebody please tell me, what i've got to change in the PHP file, so, that the databinary option used by curl is going to work?



Thanks a lot for your helping toughts!










share|improve this question



























    1















    first i've to say, that i'm totally new to cURL. But I've just code some lines with curl and PHP to Upload some XML files.
    The following Code is working very good. (just looked around and build up a simple working solution)



    Now i've got the issue to use the curl --Data-Binary option instead of the -F (Form) option.



    The following Code is in the php-File (curlupload.php):



    <?php

    @$authcode = htmlspecialchars($_POST["acode"]);

    if ($authcode == "1234"){
    echo "AuthCode okn";

    $upstatus = "1";
    $uploadpath = "files/";
    $filedata = $_FILES['file']['tmp_name'];
    $target_file = $uploadpath . basename($_FILES["file"]["name"]);

    #echo "Acode:" . $authcode . "n";
    #echo "filedata= ".$filedata."n";
    echo "target_file= ".$target_file."n";
    #echo "1 upstatus= ".$upstatus."n";

    if (strpos($target_file, ".xml") == true) {
    $upstatus = "1";
    echo "2 upstatus= ".$upstatus."n";

    if (file_exists($target_file)) {
    echo "Sorry, file already exists.n";
    $upstatus = "0";
    }

    #echo "4 upstatus= ".$upstatus."n";
    if ($upstatus == "1") {
    if ($filedata != '')
    copy($filedata,$target_file);
    echo "n-----n";
    echo "successn";
    }

    }else{
    $upstatus = "0";
    #echo "3 upstatus= ".$upstatus."n";
    echo "Filenamecheck Failed.n";
    }



    if ($upstatus == "0") {
    echo "Sorry, your file was not uploaded.n";
    }
    }else{
    echo "NO.";
    }
    ?>


    my cURL command for uploading a file is the following:



    curl -v -F 'file=@/root/testfile01.xml' -F 'acode=1234' http://webserver/upload2/curlupload.php


    Now i'm going to use the -d and --data-binary option like this:



    curl --data-binary "file=@testfile02.xml" --data "acode=1234" http://webserver/upload2/curluploadtest.php


    with the 2nd above command i receive the following messages:



    AuthCode ok
    target_file= files/
    Filenamecheck Failed.
    Sorry, your file was not uploaded.


    so, the variables got no input - Can somebody please tell me, what i've got to change in the PHP file, so, that the databinary option used by curl is going to work?



    Thanks a lot for your helping toughts!










    share|improve this question

























      1












      1








      1








      first i've to say, that i'm totally new to cURL. But I've just code some lines with curl and PHP to Upload some XML files.
      The following Code is working very good. (just looked around and build up a simple working solution)



      Now i've got the issue to use the curl --Data-Binary option instead of the -F (Form) option.



      The following Code is in the php-File (curlupload.php):



      <?php

      @$authcode = htmlspecialchars($_POST["acode"]);

      if ($authcode == "1234"){
      echo "AuthCode okn";

      $upstatus = "1";
      $uploadpath = "files/";
      $filedata = $_FILES['file']['tmp_name'];
      $target_file = $uploadpath . basename($_FILES["file"]["name"]);

      #echo "Acode:" . $authcode . "n";
      #echo "filedata= ".$filedata."n";
      echo "target_file= ".$target_file."n";
      #echo "1 upstatus= ".$upstatus."n";

      if (strpos($target_file, ".xml") == true) {
      $upstatus = "1";
      echo "2 upstatus= ".$upstatus."n";

      if (file_exists($target_file)) {
      echo "Sorry, file already exists.n";
      $upstatus = "0";
      }

      #echo "4 upstatus= ".$upstatus."n";
      if ($upstatus == "1") {
      if ($filedata != '')
      copy($filedata,$target_file);
      echo "n-----n";
      echo "successn";
      }

      }else{
      $upstatus = "0";
      #echo "3 upstatus= ".$upstatus."n";
      echo "Filenamecheck Failed.n";
      }



      if ($upstatus == "0") {
      echo "Sorry, your file was not uploaded.n";
      }
      }else{
      echo "NO.";
      }
      ?>


      my cURL command for uploading a file is the following:



      curl -v -F 'file=@/root/testfile01.xml' -F 'acode=1234' http://webserver/upload2/curlupload.php


      Now i'm going to use the -d and --data-binary option like this:



      curl --data-binary "file=@testfile02.xml" --data "acode=1234" http://webserver/upload2/curluploadtest.php


      with the 2nd above command i receive the following messages:



      AuthCode ok
      target_file= files/
      Filenamecheck Failed.
      Sorry, your file was not uploaded.


      so, the variables got no input - Can somebody please tell me, what i've got to change in the PHP file, so, that the databinary option used by curl is going to work?



      Thanks a lot for your helping toughts!










      share|improve this question














      first i've to say, that i'm totally new to cURL. But I've just code some lines with curl and PHP to Upload some XML files.
      The following Code is working very good. (just looked around and build up a simple working solution)



      Now i've got the issue to use the curl --Data-Binary option instead of the -F (Form) option.



      The following Code is in the php-File (curlupload.php):



      <?php

      @$authcode = htmlspecialchars($_POST["acode"]);

      if ($authcode == "1234"){
      echo "AuthCode okn";

      $upstatus = "1";
      $uploadpath = "files/";
      $filedata = $_FILES['file']['tmp_name'];
      $target_file = $uploadpath . basename($_FILES["file"]["name"]);

      #echo "Acode:" . $authcode . "n";
      #echo "filedata= ".$filedata."n";
      echo "target_file= ".$target_file."n";
      #echo "1 upstatus= ".$upstatus."n";

      if (strpos($target_file, ".xml") == true) {
      $upstatus = "1";
      echo "2 upstatus= ".$upstatus."n";

      if (file_exists($target_file)) {
      echo "Sorry, file already exists.n";
      $upstatus = "0";
      }

      #echo "4 upstatus= ".$upstatus."n";
      if ($upstatus == "1") {
      if ($filedata != '')
      copy($filedata,$target_file);
      echo "n-----n";
      echo "successn";
      }

      }else{
      $upstatus = "0";
      #echo "3 upstatus= ".$upstatus."n";
      echo "Filenamecheck Failed.n";
      }



      if ($upstatus == "0") {
      echo "Sorry, your file was not uploaded.n";
      }
      }else{
      echo "NO.";
      }
      ?>


      my cURL command for uploading a file is the following:



      curl -v -F 'file=@/root/testfile01.xml' -F 'acode=1234' http://webserver/upload2/curlupload.php


      Now i'm going to use the -d and --data-binary option like this:



      curl --data-binary "file=@testfile02.xml" --data "acode=1234" http://webserver/upload2/curluploadtest.php


      with the 2nd above command i receive the following messages:



      AuthCode ok
      target_file= files/
      Filenamecheck Failed.
      Sorry, your file was not uploaded.


      so, the variables got no input - Can somebody please tell me, what i've got to change in the PHP file, so, that the databinary option used by curl is going to work?



      Thanks a lot for your helping toughts!







      php curl post file-upload






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 24 '18 at 22:57









      AlexAlex

      61




      61
























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