First question:
when i want to upload a file i get the message
my source code is the following:
demo2.php:
<form method="post"> <input type="text" name="test" id="test" /> <?php $uploader=new PhpUploader();
$uploader->MultipleFilesUpload=true; $uploader->InsertText="Select multiple files (Max 10M)"; $uploader->MaxSizeKB=10240; $uploader->AllowedFileExtensions="*.jpg,*.png,*.gif,*.bmp"; $uploader->UploadUrl="demo2_upload.php"; $uploader->Render(); ?> </form> <script type='text/javascript'> function CuteWebUI_AjaxUploader_OnTaskComplete(task) { var div=document.createElement("DIV"); var link=document.createElement("A"); link.setAttribute("href","savefiles/myprefix_"+task.FileName); link.innerHTML=task.test+"You have uploaded file : savefiles/myprefix_"+task.FileName; link.target="_blank"; div.appendChild(link); document.body.appendChild(div); } </script> demo2_upload.php: $uploader=new PhpUploader();
$mvcfile=$uploader->GetValidatingFile();
if($mvcfile->FileName=="accord.bmp"){ $uploader->WriteValidationError("My custom error : Invalid file name. "); exit(200);}
//USER CODE:echo $_POST['test'];$targetfilepath= "/home/www/web2/html/cycle4/uploads/uploads/files/" . $mvcfile->FileName;if( is_file ($targetfilepath) ) unlink($targetfilepath);$mvcfile->MoveTo( $targetfilepath );
$uploader->WriteValidationOK();
Second question:
As you can see, I have in demo2.php an input field "test". How can i access the value of this field? I want to write it in sql-database . can i read the value of this field in demo2_upload.php with $_POST['test']?