javedkarim

try this out and see whats the error $result = mysql_query("SELECT table1.username, table2.username FROM table1, table2 WHERE username = '".$_POST['username']."'") or die ( mysql_error() ) ; I hope it will solve the problem

Well this script runs fine on my ip address ie if i replace my ip address with $ip = '10.0.0.3'; I think there is sort of server site restriction but in windows XP i go to start>Run>\\192.168.12.87 or any other ip. Other people sharing are easily accessed even there directories and i can easily copy the file from them. Same applies on my ip if i access my IP address using the same method it also access and other can also copy file form my ip. The problem is when i try to list the files using php from other people IP address it wont let me list them even i can access them but from my ip i can easily list the files please tell me how can i list other people directories. Is Apache has some kind of restriction or there is other problem.

Please some one help me in this matter. :'(

Thanks for the reply here is my replaced statment $ip = '192.168.62.35'; $path = '//'.$host.'/Dethklok - The Dethalbum (2007)/'; or $path = '//'.$host.'/Dethklok - The Dethalbum (2007)'; Well i have tried your script. At if ( !is_dir($path) ) die( $path. ' is not a directory' ); line it dies. here is the error message. //HYDRAZDUNGEON/Dethklok - The Dethalbum (2007) is not a directory even directory exists and directory full path is \\192.168.62.35\Dethklok - The Dethalbum (2007) Up till now i have figured that the problem is with directory path. If i am giving the wrong path correct me.

Well your script works many thanks its really cool that was the stuff i was looking for. But there is problem there The code is still showing me warning Warning: opendir(//HYDRAZDUNGEON/TV Series/House MD/Season 4) [function.opendir]: failed to open dir: No error in C:\wamp\www\tutorials\directory\dir_list_1.php on line 6 Warning: readdir(): supplied argument is not a valid Directory resource in C:\wamp\www\tutorials\directory\dir_list_1.php on line 17 Warning: closedir(): supplied argument is not a valid Directory resource in C:\wamp\www\tutorials\directory\dir_list_1.php on line 22 Well i am on local area network where every one shares there stuff. Tell me about administrative stuff if its really need to get permission i will defiantly get permission. If its not important leave administrative stuff. Because i have to build search engine i have to focus on that stuff. So please help me to sort the problem of this directory.

All above methods works perfectly with my computer drive directories but not works with Ip method( network computers ) ie. $dir = opendir('//192.168.62.35/TV Series/House MD/Season 4'); Your right about mapping the network drives but i can not map 400+ computers they are too much for network drive and how will i get all 400 computers name. $url = '//computername/path/to/file'; Tell me how i will get all computers name. Is there any other methods in php and Apache.

Thanks it worked it works $dir = opendir("C:\wamp\www\\tutorials\directory"); or $dir = opendir("C:\\wamp\\www\\tutorials\\directory"); But it did not worked with network files i tried to use it on network directories see below. Note rest code is same as below. $dir = opendir("\\192.168.62.35\TV Series\House MD\Season 4"); or $dir = opendir("\\192.168.62.35\\TV Series\\House MD\\Season 4"); It gives me error Warning: opendir(\192.168.62.35\TV Series\House MD\Season 4) [function.opendir]: failed to open dir: No error in C:\wamp\www\tutorials\directory\dir_list_1.php on line 6 Warning: readdir(): supplied argument is not a valid Directory resource in C:\wamp\www\tutorials\directory\dir_list_1.php on line 15 Warning: closedir(): supplied argument is not a valid Directory resource in C:\wamp\www\tutorials\directory\dir_list_1.php on line 20 I have used Ip address. Is it the right way to list the network directory files? or there are other ways to list the network directories. please help me.

Thanks for help I have done as you have said: <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); //Open images directory $dir = opendir("C:\wamp\www\tutorials\directory") or die ("File cant be opened"); echo "<br />"; echo "-----> $dir "; echo "<br />"; //List files in images directory while (($file = readdir($dir)) !== false) { echo "filename: " . $file . "<br />"; } closedir($dir); ?> It gives me this error Warning: opendir(C:\wamp\www utorials\directory) [function.opendir]: failed to open dir: No error in C:\wamp\www\tutorials\directory\dir_list_1.php on line 5 File cant be opened

Hi, I am using WAMP server for php and mysql and running windows XP. here is the link to wamp sever: http://www.wampserver.com/en/ I am trying to list all the files in the directory and here is my code <?php //Open images directory $dir = opendir("C:\wamp\www\tutorials\directory") or die ("File cant be opened"); echo "<br />"; echo "-----> $dir "; echo "<br />"; //List files in images directory while (($file = readdir($dir)) !== false) { echo "filename: " . $file . "<br />"; } closedir($dir); ?> When i ran the code it gives me error: File cant be opened Its really annoying. I have tried several time to use opendir() function every time i fails. please tell me what is the error.