Avatar billede smeier Nybegynder
23. august 2007 - 15:06 Der er 1 kommentar og
1 løsning

Kan man sende flere filer gennem $_files samtidigt.

Jeg finder 2 filer i min form.
<input type="file" name="satb_file" size="20">
<input type="file" name="sheet_file" size="20">

Og forsøger mere eller mindre at hente begge filer således:

if($_POST["action"] == "Upload Fil(er)")
{
unset($imagename);
unset($imagename2);


if(!isset($_FILES) && isset($HTTP_POST_FILES))
$_FILES = $HTTP_POST_FILES;


if(!isset($_FILES['sabt_file']))
$error["satb_file"] = "Du har ikke uploadet en SABT fil.";

if(!isset($_FILES['sheet_file']))
$error["sheet_file"] = "Du har ikke uploadet en SHEET fil.";


$imagename = basename($_FILES['satb_file']['name']);
$imagename2 = basename($_FILES['sheet_file']['name']);

$navn = $_POST['zipnavn'];

//echo $imagename;

if(empty($imagename))
$error["imagename"] = "The name of the image was not found.";

if(empty($imagename2))
$error["imagename2"] = "The name of the image was not found2.";

if(empty($error))
{
$newimage = "filer/" . $imagename;
$newimage2 = "filer/" . $imagename2;

//echo $newimage;
$result = @move_uploaded_file($_FILES['satb_file']['tmp_name'], $newimage);
$result2 = @move_uploaded_file($_FILES['sheet_file']['tmp_name'], $newimage2);

Kan man det?
Kan ihvertfald ikke få det til at virke.
Mvh Troels
Avatar billede smeier Nybegynder
23. august 2007 - 15:08 #1
Glemte lige slutningen af det hele.

$sabtname = $imagename;
    $sheetname = $imagename2;
   
 
  // insert values
$query = "INSERT INTO finnroar_tabel (sabtname, sheetname, navn)
VALUES('$sabtname','$sheetname','$navn')";
    $result = mysql_query($query);
      if (mysql_affected_rows() == 1) {
    echo "Filen er nu succesfuldt uploadet.";
  } else {
    error_log(mysql_error());
    echo "<P>Something went wrong.</P>";
  }
}
  else{
  // Show the form
 

if(empty($result))
$error["result"] = "There was an error moving the uploaded file.";
}
if(empty($result2))
$error["result2"] = "There was an error moving the uploaded file2.";


}

include("upload_form_slideshow.php");

if(is_array($error))
{
while(list($key, $val) = each($error))
{
echo $val;
echo "<br>\n";
}
}
Avatar billede smeier Nybegynder
23. august 2007 - 15:59 #2
løste det således:
if($_POST["action"] == "Upload Fil(er)")
{
unset($imagename);
unset($imagename2);


if(!isset($_FILES) && isset($HTTP_POST_FILES))
$_FILES = $HTTP_POST_FILES;




if(!isset($_FILES['sheet_file']))
$error["sheet_file"] = "Du har ikke uploadet en SHEET fil.";


$imagename = basename($_FILES['satb_file']['name']);
$imagename2 = basename($_FILES['sheet_file']['name']);

$navn = $_POST['zipnavn'];

//echo $imagename;

if(empty($imagename))
$error["imagename"] = "The name of the image was not found.";

if(empty($imagename2))
$error["imagename2"] = "The name of the image was not found2.";

if(empty($error) || empty($error2))

{
$newimage = "filer/" . $imagename;
$newimage2 = "filer/" . $imagename2;

//echo $newimage;
$result = @move_uploaded_file($_FILES['satb_file']['tmp_name'], $newimage);
$result2 = @move_uploaded_file($_FILES['sheet_file']['tmp_name'], $newimage2);

//Indsæt i database


    $sabtname = $imagename;
    $sheetname = $imagename2;
   
 

  // insert values
$query = "INSERT INTO finnroar_tabel (sabtname, sheetname, navn)
VALUES('$sabtname','$sheetname','$navn')";
    $result = mysql_query($query);
      if (mysql_affected_rows() == 1) {
    echo "Filen er nu succesfuldt uploadet.";
  } else {
    error_log(mysql_error());
    echo "<P>Something went wrong.</P>";
  }
}
  else{
  // Show the form
 

if(empty($result))
$error["result"] = "There was an error moving the uploaded file.";
}
if(empty($result2))
$error["result2"] = "There was an error moving the uploaded file2.";


}

include("upload_form_slideshow.php");

if(is_array($error))
{
while(list($key, $val) = each($error))
{
echo $val;
echo "<br>\n";
}
}
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