在php中打开动态文件

Question

I need to open files uploaded by users in blender (for data retrieval and processing).

So far creating $cmd and running it with exec(); has worked for testing but is useless to open files uploaded by users since the file being run is always predefined, (C:\\xampp\\htdocs\\test.new\\test.stl).

$cmd = '"C:\Users\DJ LX\Documents\blender.old\blender.exe" C:\xampp\htdocs\test.new\test.stl"';

if ((($_FILES["file"]["type"] == "application/netfabb")
|| ($_FILES["file"]["type"] == "application/sla")
|| ($_FILES["file"]["type"] == "fileapplication/sla")
|| ($_FILES["file"]["type"] == "fileapplication/vnd.ms-pki.stl")))
  {
  if ($_FILES["file"]["error"] > 0)
   {
   echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
   }
  else
  {
  echo "Upload: " . $file . "<br>";
  echo "Type: " . $_FILES["file"]["type"] . "<br>";
  echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
  echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
  exec($cmd);

I tried just using the $_FILES information but that doesn't work.

$cmd = "C:\Users\DJ LX\Documents\blender.old\blender.exe" $_FILES["file"]["type"];

and

$cmd = "C:\xampp\htdocs\test.new\upload" . $_FILES["file"]["type"];

I've also tried using fopen assigned as $openfile, then using exec($openfile), but an error appears stating that the 1st parameter in exec() needs to be a string.

$openfile = fopen($_FILES["file"]["type"], "r");

if ((($_FILES["file"]["type"] == "application/netfabb")
|| ($_FILES["file"]["type"] == "application/sla")
|| ($_FILES["file"]["type"] == "fileapplication/sla")
|| ($_FILES["file"]["type"] == "fileapplication/vnd.ms-pki.stl")))
  {
  if ($_FILES["file"]["error"] > 0)
   {
   echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
   }
  else
  {
  echo "Upload: " . $file . "<br>";
  echo "Type: " . $_FILES["file"]["type"] . "<br>";
  echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
  echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
  exec($opefile);

Anyway, I'm really confused on how I could take the file name from the file uploaded and use it to open the file. Any help would be greatly appreciated.

Answer 1

You're probably looking for $_FILES['file']['tmp_name'] .

Update

A file is uploaded with a temporary(almost random) name. You get that with $_FILES['file']['tmp_name'] . The file name of the source file is in $_FILES['file']['name'] . Your command line string needs a space between the executable and the file name and should probably look something like:

$cmd = '"C:\Users\DJ LX\Documents\blender.old\blender.exe" '
           . escapeshellarg( $_FILES["file"]["tmp_name"] );

It may be safe with or without the double quotes - check that for yourself.

Answer 2

Okay so I figured it out, thanks for the suggestion blue. After playing with the $_FILES['file']['tmp_name'], I was able to get blender to attempt opening it but since it was unrecognized, (tmp file) it didn't work.

I then reverted back to $_FILES["file"]["type"] but moved the exec($cmd) further down in the script, after the $_FILES["file"]["type"] is moved to the upload folder.