PHP File Uploader

Re: Unable to get file name of uploaded file

manpakhong — Sun, 12 Feb 2012 04:34:53 GMT

simply to put the code in the same page:

    <?php
       $uploader=new PhpUploader();

       $uploader->MultipleFilesUpload=true;
       $uploader->InsertText="Select multiple files (Max 1000M)";

       $uploader->MaxSizeKB=1024000;
       $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp,*.txt,*.zip,*.rar";

       $uploader->SaveDirectory="../pics/";

       $uploader->FlashUploadMode="Partial";

       $uploader->Render();

   ?>

<script type='text/javascript'>
   function CuteWebUI_AjaxUploader_OnTaskComplete(task)
   {
       var div=document.createElement("div");
       var table=document.createElement("table");

       var tableRow=document.createElement("tr");
       var tableData=document.createElement("td");
       var tableData1=document.createElement("td");

       var input=document.createElement("input");
       input.setAttribute("type","text");
       input.setAttribute("name", "albumPicComment");

       var link=document.createElement("a");
       link.setAttribute("href","../pics/"+task.FileName);
       link.innerHTML="You have uploaded file : ../pics/"+task.FileName;
       link.target="_blank";

       var img=document.createElement("img");
       img.setAttribute("src", "../pics/" + task.FileName);
       img.setAttribute("class", "postBackPics");

       div.appendChild(table);
       table.appendChild(tableRow);
       tableRow.appendChild(tableData);
       tableRow.appendChild(tableData1);
       tableData.appendChild(img);
       tableData1.appendChild(input);
       img.appendChild(link);
       document.body.appendChild(div);
   }
   </script>

Man Pak Hong, Dave

manpakhong@hotmail.com

Re: Unable to get file name of uploaded file

Kenneth — Mon, 26 Sep 2011 08:43:40 GMT

Hi attourath,

The line below shows you how to get the upload file name in my example

echo "myfile=" . $mvcfile->FileName;

Regards,

Ken

Re: Unable to get file name of uploaded file

attourath — Sat, 24 Sep 2011 07:08:19 GMT

I used your file :

---------

function CuteWebUI_AjaxUploader_OnTaskComplete(task)

{

var div=document.createElement("DIV");

var link=document.createElement("A");

link.setAttribute("href","admin/testUploader/myprefix_"+task.FileName);

link.innerHTML="You have uploaded file : savefiles/myprefix_"+task.FileName;

link.target="_blank";

div.appendChild(link);

document.body.appendChild(div);

}

</script>

------------

Could you tell me how can I get FileName and put it in : <?php $fileUploaded =

task.FileName; ?>

many thanks in advance

regards

Attourath

Re: Unable to get file name of uploaded file

NatalieWillson — Mon, 25 Jul 2011 04:10:01 GMT

I was looking for this. Thanks for sharing.

Re: Unable to get file name of uploaded file

Eric — Sat, 25 Jun 2011 08:04:34 GMT

Hi DBarsri,

I tried the following code, those two highlighted code are commented, this example works fine on my end, the mvcfle return correct value, please check whether those highlighted code leads to your issue:

<?php require_once "phpuploader/include_phpuploader.php" ?>
<?php session_start(); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
</head>
<body>
<div class="demo">
   <form id="form1" method="POST">
    <?php
                           $uploader=new PhpUploader();
                           $uploader->Name="current_uploader";
                           $uploader->MultipleFilesUpload=true;
                           $uploader->InsertText="Select file(s)";
                           $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp,*.txt,*.zip";
                           //$uploader->SaveDirectory="files/" . $_SESSION['orgid'];
                           $uploader->FlashUploadMode="Partial";
                           $uploader->Render();
    ?>
<?php
$fileguidlist=@$_POST["current_uploader"];
   if($fileguidlist)
   {
      //echo $fileguidlist;
      $guidlist=split("/",$fileguidlist);
      foreach($guidlist as $fileguid)
    {
       if($uploader)
          echo $fileguid;
     $mvcfile=$uploader->GetUploadedFile($fileguid);
     if($mvcfile)
     {
            echo "myfile=" . $mvcfile->FileName;
            //Rename the uploaded file
            //rename("files/".$_SESSION['orgid']."/".$mvcfile->FileName,$_SESSION['curTkt']."~".$mvcfile->FileName);
         }
      }
   }
?>
    <br /><br />
    <input type='submit' value="Submit Form" />
   </form>
</div>
</body>
</html>

Thanks for asking

Eric@cutesoft.net

Re: Unable to get file name of uploaded file

DBarsri — Fri, 24 Jun 2011 10:51:44 GMT

Here is the creation code:

                        <?php
                           $uploader=new PhpUploader();
                           $uploader->Name="current_uploader";
                           $uploader->MultipleFilesUpload=true;
                           $uploader->InsertText="Select file(s)";
                           //$uploader->MaxSizeKB=1024000;
                           $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp,*.txt,*.zip";
                           $uploader->SaveDirectory="files/" . $_SESSION['orgid'];
                           $uploader->FlashUploadMode="Partial";
                           $uploader->Render();
                        ?>

Thanks

Re: Unable to get file name of uploaded file

Eric — Thu, 23 Jun 2011 15:56:59 GMT

Hi DBarsri,

Can you post the creation code of php uploader here?

It should look like the following:

<?php
    $uploader=new PhpUploader();
    $uploader->MaxSizeKB=10240;
    $uploader->MultipleFilesUpload=true;
    $uploader->Name="myuploader";
    $uploader->InsertText="Select multiple files (Max 10M)";
    $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp,*.txt,*.zip,*.rar";
    $uploader->Render();
?>

Thanks for asking

Eric@cutesoft.net

Re: Unable to get file name of uploaded file

DBarsri — Thu, 23 Jun 2011 13:07:18 GMT

How would I check if the guid is correct?

Thanks

EDIT - I checked the posted guid with the js task guid and they match. I cannot seem to find the difference between my code and the form-keepingstate.php example. My source code has a lot of dependencies and it is part of a larger web application with many pages and built-in security so posting the full code is probably not a good idea. Here is what it looks like right now:

   $fileguidlist=@$_POST["current_uploader"];
   if($fileguidlist)
   {
      //echo $fileguidlist;
      $guidlist=split("/",$fileguidlist);
      foreach($guidlist as $fileguid)
   {
       if($uploader)
          echo $fileguid;
     $mvcfile=$uploader->GetUploadedFile($fileguid);
     if($mvcfile)
     {
            echo "myfile=" . $mvcfile->FileName;
            //Rename the uploaded file
            rename("files/".$_SESSION['orgid']."/".$mvcfile->FileName,$_SESSION['curTkt']."~".$mvcfile->FileName);
         }
      }
   }

It is still not getting into the if($mvcfile) block

Thanks

Re: Unable to get file name of uploaded file

Eric — Thu, 23 Jun 2011 12:22:27 GMT

Did you check whether $fileguid and $_POST["current_uploader"] have correct value?

You can post your full code here, we will run your code and check it for you.

Thanks for asking

eric@cutesoft.net

Re: Unable to get file name of uploaded file

DBarsri — Thu, 23 Jun 2011 09:55:51 GMT

My problem isn't so much getting a filename, it is getting the object that will give me the filename. I changed my code to match the example but am still getting the same result. Here is the new code:

   $fileguidlist=@$_POST["current_uploader"];
   if($fileguidlist)
   {
      echo $fileguidlist;
      $guidlist=split("/",$fileguidlist);
      foreach($guidlist as $fileguid)
   {
     $mvcfile=$uploader->GetUploadedFile($fileguid);
     if($mvcfile)
     {

// Never gets here

}

I'm wondering if $uploader is losing the uploaded files after the post although earlier I tried making it a session variable and that didn't work either.

Re: Unable to get file name of uploaded file

Kenneth — Wed, 22 Jun 2011 22:09:30 GMT

Hi DBarsri,

Please check the code of example "form-keepingstate.php", it shows you how to get the file name by property "FileName".

Online demo http://phpfileuploader.com/demo/form-keepingstate.php

Regards,

Ken

Unable to get file name of uploaded file

DBarsri — Wed, 22 Jun 2011 17:07:08 GMT

I am trying to rename a file after it has been uploaded but I am unable to get the file name. I can get the GUID for the file but not the file name. Any ideas why $uploader is not returning a phpuploadfile object?

Here is how I instantiate the uploader:

                  <form id="form1" method="POST">
                        <?php
                           $uploader=new PhpUploader();
                           $uploader->Name="current_uploader";
                           $uploader->MultipleFilesUpload=true;
                           $uploader->InsertText="Select file(s)";
                           //$uploader->MaxSizeKB=1024000;
                           $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp,*.txt,*.zip";
                           $uploader->SaveDirectory="files/" . $_SESSION['orgid'];
                           $uploader->FlashUploadMode="Partial";
                           $uploader->Render();
                        ?>
                        </form>

Here is where I try to rename the file:

   $fileguid=$_POST["current_uploader"];
   if($fileguid)
   {
      echo $fileguid;
      $uploadedfile=$uploader->GetUploadedFile($fileguid);
      echo "myfile=" . $uploadedfile->FileName;
      if($uploadedfile)
      {
         //Rename the uploaded file
         echo "Renaming file";
         $uploadedfile->MoveTo("./".$_SESSION['curTkt']."-".$uploadedfile->FileName);
      }
   }

The echoes are for debugging. It echoes the fileguid but then $uploadedfile->FileName is empty. It never gets to the part that echoes "Renaming file."