[RESOLVED] count the .pdf in certain directory

[blackhorse]

$path=$_SERVER['DOCUMENT_ROOT']."/pdf/";

	if (is_dir($path))
	{
		chdir($path);	
		$dir=opendir(".");
		while ($file=readdir($dir))	
		{
			//if (($file!=".")&&($file!=".."))
                                                            if (substr($file, -4)==".pdf")
			{
				$pdf_count++;
								}

		}

I am trying to find out how many pdf in a directory such as in directory "/pdf/", is this the right way? any codes more clear and brief? Thanks!

[bpat1434]

function countFiles($path='.', $extension='.pdf')
{
  $pdfs = 0;
  if(!is_dir($path))
  {
    return FALSE;  // Not a real directory, fail
  }
  else
  {
    // Proper directory, now let's search:
    $path = (substr($path, -1)=='/')?substr($path, 0, -1):$path; // remove trailing slash
    $d = dir($path);
    while( ($files = $d->read()) != FALSE)
    {
      if(!is_dir($path.'/'.$files) && ($files != '.' && $files != '..') && !is_link($path.'/'.$files))
      {
        // Not a directory, not "current" or "parent" directory links, not a symbolic link
        if(strrchr($files, '.') == $extension)
        {
          // We've found a PDF file!!! whew!!
          $pdfs++;
        }
      }
    }
    return $pdfs;
  }
}

It's a little more drawn out, but does what you have, and adds a few more checks.

To optimize your code, I'd say that you use [man]strrchr/man instead of substr() since [man]strrchr[/man] will find the last occurance of a character in a string. So it essentially searches from the right for the character you specify. It then returns everything from that point on (giving you only the extension). It's a little faster than the substr() call.

Now, to call this for the current directory, just do:

$num_of_pdfs = countFiles();

If there are any PDFs in that directory, it will count them. If not, it will return 0. If it's not actually a directory, or can't get to the directory (like paths that require apache's mod_rewrite) then it will return false. Obviously to check that it returned something you'd do:

$num_of_pdfs = countFiles('.', '.pdf');  // equivalent to countFiles();
if($num_of_pdfs === FALSE) // Check type of var (bool) & value (FALSE)
// 0 is an integer, and won't cause this to go off
{
  echo 'Sorry, the directory couldn\'t be searched';
}
else
{
  echo $num_of_pdfs.' were found!!';
}

[BobLennon]

You almost had it right!

$path=$_SERVER['DOCUMENT_ROOT']."/pdf/";
if (is_dir($path)){
$dir=opendir($path);
while ($file=readdir($dir)) {
if (substr($file, -4)==".pdf")$pdf_count++;
}
}

[bpat1434]

You still have a mistake Bob... where is $pdf_count initialized? It's not in your code. So php would return an error.... can't increment what's not there....

And if he left it off 'cuz he initialized it elsewhere, then he needs to specify that, rather than having us in the unknown...

[BobLennon]

bpat1434 "You still have a mistake Bob... where is $pdf_count initialized"

Sorry to disagree bpat1434, but the The code is correct. Try it!

"So php would return an error..

It does not because there is no need to initialize php variables in cases like this. A variable whose value has not been set behaves like the null value. Check it out in the manual.

[bpat1434]

I regret to inform you that we're both correct:

Notice: Undefined variable: k in I:\Websites\count.php on line 7
50

<?php

for($i=0; $i<100; $i++)
{
	if($i%2==0)
	{
		$k++;
	}
}
echo $k;

?>

Even though it's just a warning, it's still an error that needs to be supressed, or taken care of with just a simple $k=0..... (or in this case $pdf_count = 0)...

[BobLennon]

I regret to inform you that we're both correct:

Quote:
Notice: Undefined variable: k in I:\Websites\count.php on line 7
50

PHP Code:
<?php

for($i=0; $i<100; $i++)
{
if($i%2==0)
{
$k++;
}
}
echo $k;

?>

Even though it's just a warning, it's still an error

Not to cause a problem bpat1434, but your statement

"Even though it's just a warning, it's still an error " is, in itself, a contradiction.

And certainly we both cant be correct when you say:

The code submitted produces no errors. A warning is not an error.

[bpat1434]

Okay, fine how about this:

There is still inheritly wrong with the code because rather than parsing perfectly fine without any errors or warnings, one is produced due to the fact that a variable isn't instantiated, in one form or another, and thus an "ugly" or needless message is now shown on the page and possibly giving information about the server away.

Is that acceptable? Oh, and while I'm thinking about it... why is it in php.ini called E_WARNING.... E standing for an ERROR of LEVEL WARNING... it's an ERROR LEVEL... so it's still an error....

PHP Manual wrote:

To report errors from an internal function, you should call the php3_error() function. This takes at least two parameters -- the first is the level of the error, the second is the format string for the error message (as in a standard printf() call), and any following arguments are the parameters for the format string. The error levels are:
E_NOTICE
E_WARNING
E_ERROR
E_PARSE
E_CORE_ERROR
E_CORE_WARNING
E_COMPILE_ERROR
etc... etc...

[BobLennon]

I regret to inform you that we're both correct:...Even though it's just a warning, it's still an error

Not to cause a problem bpat1434, but your statement

"Even though it's just a warning, it's still an error " is, in itself, a contradiction.

And certainly the script does exactly what is intended.

It was tested with the following statements in effect:

ini_set('display_errors',1);
error_reporting(E_ALL & ~E_NOTICE);

Without any errors.

However, your citing other code that produces a warning under different circumstances is out of context.

And why would one regret that we are both correct?

Bottom line is that we both tried to help, and like the Perl folks say, "There's More Than One Way To Do It."

Best regards...BobLennon

[bpat1434]

The point of my code was to test your theory that if you increment a non-instantiated variable, you'd get an error (what level it didn't matter, point was it generates an error). So I created pseudo code that looped and mimicked (by incrementing) a variable not yet called. Whether I loop through a directory or not is a moot point. I'm in a loop incrementing a variable I hadn't instantiated. That was the scenario.

And your error level is not what most servers have it set at. I've yet to see a server not show me a parse error. Which would imply that they've got a level similar to:
error_reporting = E_ALL | E_STRICT
NOTE: I'm using php 5

Whatever, I'm done. We were both right. Period. End of discussion.

[Weedpacket]

Um,

count(glob($_SERVER['DOCUMENT_ROOT']."/pdf/*.pdf"));

? Assuming there are no directories in there with names ending with ".pdf", of course.

[BobLennon]

Gorgeous. Elegant....