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$file['type']


geoffl1

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how are you assigning $file...show a little more code
I'm assuiming it's an upload file array
but I can't be too sure

and if that is the case
remember you have to include the name of the file upload input box from you HTML

E.G
if you're box looks like this
[code]<input type="file" name="pictures[]" />[/code]

you're print statement must be
[code]print $_FILES['pictures']['type'];[/code]
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I'm reading files from a directory. Here is the code. Thanks again.


// open the current directory by opendir
$handle=opendir(".");

while (($file = readdir($handle))!==false) {

    print $file['type'];

if($file['type']=="application/x-php"||$file['type']=="text/html"){
                //doesn't allow .php or .html files to be seen
        }
else{
print "<a href='$file'>Click to download </a>";
print $file.'<br>';
}
}
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the reason it doesn't work is because readdir() doesn't return an array
it return a string with JUST the filename in it

you'll have to use the mime_content_type() function to check the filetype

as simple change to your script
[code]
while (($file = readdir($handle))!==false) {

  if(mime_content_type ($file)=="application/x-php"| | mime_content_type ($file)=="text/html"){
               //doesn't allow .php or .html files to be seen
       }
  else{
     print "<a href='$file'>Click to download [/url]";
     print $file.'
';
  }
}

[/code]
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I don't know what you did either
readdir() when used right.....will give you the filename...and that's it

$file['type'] is kinda....undefined

when you say it's not printing anything........are you testing this script on a php file/html file
because from you included script....it's not doing anything...explaining the blank print
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