Dear Atkinson ,
Thanks you for all your help , You are the best for ever I saw in the net , who can understand well and suggestion good . Normally people in the net just try to showoff but they never wanted to tell right answer even if they know , But you help me very nicely . Thanks again , Also I sent the final code which i made with your help may be useful for some body else who is use this kind of cameras .
_________________________________________
<?php
$uploaddir = $_SERVER["DOCUMENT_ROOT"] . '/test/live/';
$headers = apache_request_headers();
$strData = $headers['Content-Disposition'];
$arrayFilename = explode('"', $strData) ;
$myFile =$arrayFilename[1];
$fh = fopen($uploaddir.$myFile, 'w') or die("can't open file");
$data = file_get_contents("php://input");
fwrite($fh, "$data\t\n");
fclose($fh);
?>
_________________________________________
________________________________
From: "Atkinson, Robert" <ratkinson@xxxxxxxxxxxxx>
To: php-objects@xxxxxxxxxxxxxxx
Sent: Tuesday, November 25, 2008 4:52:13 PM
Subject: RE: Converting CGI+Perl to PHP Help
This should do it (untested) :-
$stringData = $headers['Content- Disposition' ];
$arrayFilename = explode("=", $stringData) ;
$imageName = replace("\"" ,"",$arrayFilena me[1]);
-----Original Message-----
From: php-objects@ yahoogroups. com [mailto:php-objects@ yahoogroups. com] On
Behalf Of Ali Jamali
Sent: 25 November 2008 12:37
To: php-objects@ yahoogroups. com
Subject: Re: Converting CGI+Perl to PHP Help
I need the file name because the camera give images name by different setting
, For example user wanted add date & time to file or some specific name they
declare and image must same in the server with that . Any way I try to do
that
I use this line
$stringData = $headers['Content- Disposition' ];
and out put is :
attachment; filename="image. jpg"
So Is there any way I just take the filename or not ?
I just need image.jpg
I will become very perfect if you suggestion something to fetch the name .
Thanks
Ali
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