Link to open newest file in drectory

I'd do something like this for function getNewest();

Sorry if its a bit hokey, I slapped it together in a hurry because I have to go somewhere - if it doesn't work, I'll help you fix.

$handle=opendir($mydir);
while (false!==($filename=readdir($handle))) {
  if ( ($filename!="..")&&($filename!=".") ) {
    $stamp=filemtime("$mydir/$filename");
    array_push($files,"$stamp:$filename");
  }
}
sort($files);
$latest=$files[0];
$parts=explode(":",$latest);
$filename=$parts[0];
return($filename);

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ugh should be $filename=$parts[1]; for line 11.

sorry my php is pretty basic what would the whole code look like
in a pinch and need it up before the am

Replace your function with mine.

function get_newest($directory) {
  $handle=opendir($mydir);
  while (false!==($filename=readdir($handle))) {
    if ( ($filename!="..")&&($filename!=".") ) {
      $stamp=filemtime("$mydir/$filename");
      array_push($files,"$stamp:$filename");
    }
  }
  sort($files);
  $latest=$files[0];
  $parts=explode(":",$latest);
  $filename=$parts[1];
  return($filename);
}

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ok i will give this a try. on a side note i am linking to a php file that should open the latest xls correct.

and should this php file be in the same as the folder where the xls files are uploaded? and do i still need this code

$outfile = "RateSheets/".date("m-d-Y His").".xls";
$path = "RateSheets/";

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I made a mess out of that.  Here is one that I just tested, and it works beautifully.

<?php

$directory="/path/to/somewhere";
$filename=get_newest($directory);
echo "Newest file in $directory is $filename\n";

function get_newest($directory) {
  $files=array();
  $handle=opendir($directory);
  while (false!==($filename=readdir($handle))) {
    if ( ($filename!="..")&&($filename!=".") ) {
      $stamp=filemtime("$directory/$filename");
      array_push($files,"$stamp:$filename");
    }
  }
  rsort($files);
  $latest=$files[0];
  $parts=explode(":",$latest);
  $filename=$parts[1];
  return($filename);
}

?>

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thank you i will try it.

is this php file in the same folder as the uploaded files or a different location, i tried it no xls file pops up to open or save

No, this php file goes wherever you want in your web tree.  Your question never mentioned popping up any files to save - you just said you wanted to open it.

Ultimately, once you've identified the name of the newest file that you want to serve up the web site visitor, you have some options:

1)  If the file is already in your web root somewhere, you can just do a hyperlink:
echo "<a href=filedir/$filename>Click here to download $filename</a>\n";

2)  You can fopen() the file on the server and parse the binary data into a variable, then you can echo the appropriate headers to the browser which will prompt the user to save or open.  #1 above is easier, but if your file directory isn't visible in your web server root, the latter is what you'll have to do.  If that is the case, let me know and I'll give you the code.

i am getting this error when i go to www.afr-wholesale.com/rates.php

Warning: opendir(/RateSheets/) [function.opendir]: failed to open dir: No such file or directory in /home/content/j/o/e/joesimmel/html/rates.php on line 9

Warning: readdir(): supplied argument is not a valid Directory resource in /home/content/j/o/e/joesimmel/html/rates.php on line 10
Newest file in /RateSheets/ is

i basically want to do a link that opens the latest file.

the rates.php is the code you helped me create which queries the directory for the newest file and should open it

so in a synopsis i would like to go to www.afr-wholesale.com/rates.php and the newwest file should open just as if i was going to www.afr-wholesale.com/ratesheets/name.xls

thanks, still getting this not sure why should't it just be /ratesheets ? or is it an absolute path?
/home/content/j/o/e/joesimmel/html/RateSheets
which i just tried and didnt work


Warning: opendir(/ratesheets) [function.opendir]: failed to open dir: No such file or directory in /home/content/j/o/e/joesimmel/html/rates.php on line 10

Warning: readdir(): supplied argument is not a valid Directory resource in /home/content/j/o/e/joesimmel/html/rates.php on line 11

Warning: Cannot modify header information - headers already sent by (output started at /home/content/j/o/e/joesimmel/html/rates.php:10) in /home/content/j/o/e/joesimmel/html/rates.php on line 6

actually figured that out but all gibberish comes up now

nevermind looks like i got it to go

you are awesome thank you so much

No problem, glad it does what you need!

You the man!