rnb_cassanova

Hi lads, qucik question, one command im not familar with is - $TofD = $_POST["TofD"]; can some explain? Thanks David

I have downloaded that open source calendar, but ill be honest, I have no clue where to start when installing it. Sorry, just quite new to this all. Any assistance would be helpful. Cheers lads, David

Hi Guys, I need a calendar in PHP that will allow me to mark of dates. so for example if a week is booked, i can colour code it so show this. Any ideas how I can implement this? Thanks for you help in advance, David

hi im sorry, it worked perfect my fault. thanks for all your help Ray. Solved. Cheers David

hi, i inserted that to recieve the error- <script src="http://www.google-analytics.com/urchin.js" type="text/javascript"> </script> <script type="text/javascript"> _uacct = "UA-3803417-1"; urchinTracker(); </script> im a bit lost now. if you can help i would appreciate it? Cheers David

no cigar mate, im getting this error... http://i32.photobucket.com/albums/d36/B_318is/error-upload.jpg and on IE the link is dead. Cheers david

the one you sent me the - <?php $dir = "C:/wamp/www/DV8 System/uploads"; if ($handle = opendir($dir)) { /* This is the correct way to loop over the directory. */ while (false !== ($file = readdir($handle))) { if($file != ".." && $file != "."){ echo '<a href="'.$dir."/".$file.'" target="_blank" border=0>'.$file.''; } } } ?>

it will be image files, so JPEG, BMP, GIF and PNG's i believe. any ideas? cheers david

sorry forgot to say, also if i 'save target as' it will not download also. Cheers david

fantastic, it reads from the file correctly but when i click the link, I am not given the option of saving the file but instead an error that says Firefox cannot open the file. How can i rectify this, thanks for your help lads. David

Hi, im currently getting the following error Warning: opendir(/Uploads) [function.opendir]: failed to open dir: No error in C:\wamp\www\DV8 System\TargetUploadsPage.php on line 3 I have changed the target directory, but am leaving something out. the file is within the root and called \uploads Cheers lads david

i have a target folder where the users uploads defaults to but now i want to be able to see whats in that folder, but only that folder, so in essence the best way i can describe is something like this... http://esupport.epson-europe.com/ProductHome.aspx?lng=en-GB&data=dkmEK02SU002FeroarA05EEv55UIF0kgE0cfqxevtXS8ndgU003D&tc=6#7 all of the files in one place, constantly updated and the user can download. I hope im making sense, Cheers David

hi guys, As you may have seen I have completed the upload facility, i now need to be able to retrieve them from the destination file through the system. Can anyone help with a method? Thanks David

Dennis, That worked perfect, thanks so much for that mate. Really appreciate the help. All the best, David

Hi sorry, DV8 System/Uploads/Dave.doc Warning: move_uploaded_file(DV8 System/Uploads/Dave.doc) [function.move-uploaded-file]: failed to open stream: No such file or directory in C:\wamp\www\DV8 System\ValidatingUploadAdminForm.php on line 9 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\wamp\tmp\phpE8A3.tmp' to 'DV8 System/Uploads/Dave.doc' in C:\wamp\www\DV8 System\ValidatingUploadAdminForm.php on line 9 Possible file upload attack! Here is some more debugging info:Array ( [userfile] => Array ( [name] => Dave.doc [type] => application/msword [tmp_name] => C:\wamp\tmp\phpE8A3.tmp [error] => 0 => 28160 ) ) the code is [code <!-- The data encoding type, enctype, MUST be specified as below --> <form enctype="multipart/form-data" action="ValidatingUploadAdminForm.php" method="POST"> <!-- MAX_FILE_SIZE must precede the file input field --> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> <!-- Name of input element determines name in $_FILES array --> Send this file: <input name="userfile" type="file" /> <input type="submit" value="Send File" /> </form> ] and <?php // In PHP versions earlier than 4.1.0, $HTTP_POST_FILES should be used instead // of $_FILES. $uploaddir = 'DV8 System/Uploads/'; $uploadfile = $uploaddir . basename($_FILES['userfile']['name']); echo $uploadfile; echo '<pre>'; if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) { echo "File is valid, and was successfully uploaded.\n"; } else { echo "Possible file upload attack!\n"; } echo 'Here is some more debugging info:'; print_r($_FILES); print "</pre>"; ?> sorry its everywhere, if there is a better method of pasting this please let me know. Cheers David

I have looked at yours, your is writing to a DB, mine just uploads to a local folder, thats why im confused by the why im getting so many errors. if anyone can help i would appreciate. Cheers

Hi lads, I am trying to create an upload script. I am using http://uk3.php.net/features.file-upload for advice; Although i am receiving errors. My code is as follows- $uploaddir = '/DV8 System/Uploads/'; // im not sure if this is where the problem is - am i putting the directory in correctly? $uploadfile = $uploaddir . basename($_FILES['userfile']['name']); echo $uploadfile; echo '<pre>'; <form enctype="multipart/form-data" action="ValidatingUploadAdminForm.php" method="POST"> // also should this point to a form of to the upload directory? Cheers in advance lads David

brilliant! ill give that a go in a bit, thanks mate.

Hi Guys, I need to change the status for a record in my database - In this example its from pending to approved or declined. I don't know how to go about this so any help would be greatly appreciated. Thanks Everyone.

ok but that reads the 'login' and not the 'Name' field, here is a pic of my Db- http://i32.photobucket.com/albums/d36/B_318is/db.jpg Do i now have to ecco the session within the page i want it to display within? Cheers lads

ok cool, thats created, now i want the records only with the 'Name' field that correlates to them when they login to be displayed. eg - If the login is 'Alex' i want only the records in the DB which under the field 'Name' display 'Alex' to be shown, and all the others by other users to be hidden. Thanks lads

my login page - <?php session_start(); $_SESSION['Name'] = $Name; ?> <script language="JavaScript" type="text/JavaScript"> form action="logincheck2.php" method="get"> <p align="center"> </p> <p align="center"> </p> <p align="center"><span class="style1">User Name_</span><br> <input name="Login" type="text"> my login check - <?php session_start(); $_SESSION['Name'] = $Name; require "connect.php"; $Login = $_GET['Login']; $Password = $_GET['Password']; $query = "select * from client where Login = '".$Login."' AND Password = '".$Password."'"; $result = mysql_query($query, $connection) or die ("unable to perform query<br>$query"); $row= mysql_fetch_array($result); if ($row != null) { $_SESSION['Login'] = $row['Password']; $_SESSION['Name'] = $row['Name']; header("Location: welcompage2.php"); exit(); } else { header("Location: login2.php"); exit(); } ?> Sorry for a lot of code lads, just thought it may give a better indication than what I can explain. Thanks again

I have put - <?php session_start(); $SESSION['Name'] = $Name; require "connect.php"; on my login pages and - <?php session_start(); $SESSION['Name']; require "connect.php"; Where i want to call it into, at current its not working, can anyone help where i am going wrong? Thanks Lads

ok but im still a bit lost on how to do this, do i create another -if (isset($_SESSION['Login']) == false)? but specific to the user ID? Thanks Lads.

Hi everyone, I need to display records in a database that are only specific to the user ID that is logged in. Do i need to pass through variables in a session? Currently i have - code: <?php session_start(); require "connect.php"; if (isset($_SESSION['Login']) == false){ header("Location: Login.php"); exit(); } //$query = "select * from proposals"; $query = "SELECT * FROM proposals ORDER BY proposalID DESC"; $result = mysql_query($query, $connection) or die ("mySQL Error: ".mysql_error()); ?> Thanks guys. David