Pass db variable to opendir

[jonathonoss]

I am having trouble taking a variable obtained from SQL2000, and using it as part of the directory path for an opendir() statement.

Here is the code:
while(odbc_fetch_row($cmservicelistr))
{
$cmnumber = odbc_result($cmservicelistr, 1);
$medreccountq = "select count([logmemo]) from caselogdata where (([clogcase] = $cmnumber) and (medreclink is not null))";
$medreccountr = odbc_exec($connect, $medreccountq);
$medreccount[$lt][0] = odbc_result($medreccountr,1);
$medreccount[$lt][1] = $cmnumber;

/////////////////////////////////////////////////////////
//Check to see how may CM files are in the File Directory
/////////////////////////////////////////////////////////
// $dirsearch = "cm/" . intval($cmnumber) . "/";
// $dirsearch = "cm/" . strval($cmnumber) . "/";
// $dirsearch = "cm/" . $cmnumber . "/";
// $dirsearch = "cm/$cmnumber/";
// $dirsearch = ("cm/" . $medreccount[$lt][1] . "/");
$dirsearch = ("cm/" . 41659 . "/");
echo $cmnumber . "   ";
//Identify files in PATH provided by $dirsearch
$dh = opendir($dirsearch);
while (false !== ($filename = readdir($dh))) {
$files[] = $filename;}

//Track number of files per CM#
$filescountarray[] = count($files)-2;
unset($files);
$lt++;
}

The code as shown here works. I have a loop. It gets each CM# in a query (not shown) by "$cmnumber = odbc_result($cmservicelistr, 1);" Then I take each CM# in the loop, and use it as part of a directory path to total the number of files in the directory with "$filescountarray[] = count($files)-2;".

The problem is how I take the CM# and use it as part of the directory path, as in "$dh = opendir($dirsearch);". The current code use for $dirsearch = '$dirsearch = ("cm/" . 41659 . "/");' works because I plug in 41659 (one of the CM#'s). But if I use one of the other "$dirsearch =" lines to incorporate the $cmnumber variable, it fails. Yet the line 'echo $cmnumber . "   ";' prints each cmnumber correctly. And the lines earlier that use $cmnumber, such as "$medreccountq = ..." and "$medreccount[$lt][1] = $cmnumber;" work fine. Why does $cmnumber work fine for the other statements, but not opendir()?

The error I get is not even a statement telling me which line has the error, like a typical PHP error. I either get the "Page can not be displayed" HTTP error. Or I get the "Internet Explorer cannot open the Internet site

http://..."

then below "The server returned an invalid or unrecognized response." In the IIS log, the IIS server issues a status of 200 by each request of this page, not indicating a problem at that end.

I installed SQL 2000 on a separate W2K Server box and installed PHP using PHP.EXE to parse .PHP extensions, to see if the error would go away on a different machine, but it does not.

Is this a PHP coding error?

Thanks,

Jon

 

[DRJ478]

Have you echoed out:

    $dirsearch = "cm/" . intval($cmnumber) . "/";
    $dirsearch = "cm/" . strval($cmnumber) . "/";
    $dirsearch = "cm/" . $cmnumber . "/";
    $dirsearch = "cm/$cmnumber/";

This has probably absolutely nothing to do with opendir() but the way you construct the path variable. Let's see what the echo's show.

 

[sleipnir214]

It may also have something to do with the fact that opendir() is not constrained to the current web site's document root. PHP's filesystem functions operate on the server's entire filesystem.


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[jonathonoss]

Thanks for the responses.

I just found out what it was. I am working on this alone, and sometimes just bouncing ideas/questions of others and getting their responses seems to get you to think on the right track. I thought I had taken care of this possibility yesterday when I keyed in different values for $dirsearch, as in...

'$dirsearch = ("cm/20555/");'

It was not working for the simple fact that a directory did not exist for one of the numbers!!!!!! You probably wonder how I could not have checked that before... but I already had gone through that yesterday.

True, I knew one of the directories was missing, but I was changing values for $dirsearch, putting in paths I knew didn't exist, and I was getting counts for the files in the root of the virtual directory -- or at least that was the process I had in my head. I know I was getting counts for the files in the root of the virtual directory. I am trying to follow my footsteps to see how I got the counts for the files in the root directory, but now I can't seem to recreate how I did that!

Sorry,

Jon