linkcool

well i want to transfer all the $_POST of the page because i make an uploadverify.php so the person can see what is the picture and to verify the stuff. and then i want to use the $_POST[] in uploadverifyaction.php(after the person clicks yer or no in uploadverify.php) to store everything in the database.

is it really the only way?

When we make a form, we use an action file. The $_POST[] is used there(in the "action file/page", but can we use them in the page right after the "action file"? how?

ok so i don't need to use session_is_registered(); and session_register(); anymore? (but anyway, i never really understanded what did session_register();...)

Woaw, i never tought about that but your first paragraph sayd it all, I accidently overwited the value, i writed: if($_POST['something1'] = "" ){ instead of if($_POST['something1'] == ""){ But what should i use instead of all that session stuff? and, are you sure the exit; statement will "play" if i redirect, and what it does?? Anyway, thank you very much. linkcool

Hi everybody, im linkcool, im new here and i have a little problem. I've been making some php for about 2 weeks now and i got a problem with a form and a $_POST. here is the upload.php file, it gets redirected into uploadverify.php <?php session_start(); if(!session_is_registered(username)) { session_register("wrongpage"); $_SESSION['wrongpage'] = "Sorry you are not logged in so you can't visit this page: Upload.php <br/>"; header("location:login.php"); } else{} echo '<table border="0"><tr><td width="500"> </td><td><a href="logout.php">Logout!</a></td></t></table><br/ >'; echo "All kind of files are supported <br/><br/>"; ?> <table> <form action="UploadVerify.php" method="post" enctype="multipart/form-data"> <tr> <label for="file"><td width="200">Filename:</td></label> <td><input type="file" name="file" id="file" /></td> </tr> <tr> <td>Description: (optionnal)</td><td><input type="text" name="description" /></td> </tr> <tr> <td> Album: </td> <td> <input type="text" name="something1" /> </td> </tr> <tr> <td></td><td><input type="submit" name="submit" value="Upload!" /></td> </tr> </form> </table> <?php echo 'New feature!:<br/>'; echo '<a href="view.php">View all YOUR uploaded images</a>' ?> It is a form where the user uploads a file and write something in the description textbox and something in the "something1" textbox(it was "album" before). So when the user click: upload!, it makes the uploadverify.php action. In there, the user can see the description as he writen it in the textbox, but the album dont show up. And they are both same input. I tryed the var_dump($_POST['something1']) and it does not return it as NULL, it returns as: string(0) "" . So does anyone know how to make the $_POST['something1'] have it's value? because it returns empty... And why the $_POST['description'] does not return empty? thank you, linkcool