upload script

9 January 2004, 21:46

Citaat:

Originally posted by RandyJC

PHP Code:

<html> <head> <title>Nieuwe pagina 1</title> </head> <body> <? if ($_POST['action'] != 'upit') { // hier de html voor het formulier ?> <form name=fileform method=post action=test.php enctype='multipart/form-data'> <input type='file' name='userfile'> <input type="submit" value="Verzenden" name="B1"></p> <input type='hidden' name='action' value='upit'> </form> <? $upload_mode = (@ini_get('open_basedir') || @ini_get('safe_mode')) ? 'move' : 'copy'; $source_filename = $_FILES['userfile']['name']; $destination_filename = "/home/test/public_html/img/". $_FILES['userfile']['name']; switch ($upload_mode) { case 'copy': if ( !@copy($source_filename, $destination_filename) ) { if ( !@move_uploaded_file($source_filename, $destination_filename) ) { echo "er gaat toch iets mis.<br>". __LINE__; exit; } } @chmod('./' . $source_filename, 0666); break; case 'move': if ( !@move_uploaded_file($source_filename, $destination_filename) ) { if ( !@copy($source_file, $destination_filename) ) { echo "er gaat toch iets mis.<br>". __LINE__; exit; } } @chmod($config['upload_dir'] . '/' . $destination_filename, 0666); break; } echo "Pakspul heerscht"; exit; ?> </body> </html>

krijg nu deze fout.

Parse error: parse error in /home/virtual/site82/fst/home/test/public_html/test.php on line 56
maar als ik die exit weghaal dan nog steeds

je opent nou wel een if-statement:
if ($_POST['action'] != 'upit') {

maar je sluit hm zo te zien nooit meer

dus ergens moet je nog een '}' invoegen. die plek is denk ik hier:
</form>
<?
}
$upload_mode = (@ini_get('open_basedir') || @ini_get('safe_mode')) ? 'move' : 'copy';