PHP URL variables and if statement

[i love my vans]

Hey Everyone! ( <-- First Post, Wow)

Well i want use a URL variable like this...

[PHP]www.site.co.uk/index.php?page="about"[/PHP]

With me so far? Ok.

Now when the user enters the index.php page, i want specific content loaded into a preset table, in the case above it would be the about page, but if the link were to be www.site.co.uk/index.php?page="download" the download page would load.

I already have this code snippet to display and external PHP file...

[PHP]<?php include ("http://www.site.co.uk/content/about.php"); ?>[/PHP]

But i want to be able to change the end bit from /about.php, so it would be like this....

www.site.co.uk/content + insert url variable here + .PHP

See what i mean.
Any help would be greatly appreciated.

[Troy]

http://somedomain.com/index.php?page=download

index.php:
[PHP]
<?php

if (!isset($_GET['page']) {
$page = "about"; // Default page
} else {
$page = $_GET['page'];
}

include ("/content/".$page.".php");

?>
[/PHP]

[i love my vans]

Cheers buddy, looks good, am about to go out so will test as soon as i get back in.

Cheers, ILMV

[val542]

You should put:
[PHP]
<?php

if (!isset($_GET['page']) {
$page = "about"; // Default page
} else {
$page = $_GET['page'];
}

include ("content/".$page.".php");

?>
[/PHP]

remove the slash '/' before content!!!

[Troy]

val542 is correct in that most likely, your webroot is not the root of the server, so in the example code I posted, I should have left the beginning slash off. However, the code I posted is not inaccurate--it is legal code. That is, you could make the root of your server be your DocumentRoot--in which case "/content/" may be a valid path.

val542, a little more thorough reply would be more helpful.

[i love my vans]

Um, error....

Parse error: parse error, unexpected '{' in e:\domains\m\meltdownsoftware.co.uk\user\htdocs\test web\about\index.php on line 63

Line 63 of my code is arrowed below

[PHP]<?php

if (!isset($_GET['page']) { <---------------
$page = "download"; // Default page
} else {
$page = $_GET['page'];
}

include ("content/".$page.".php");

?>[/PHP]

Any ideas?

[Troy]

If the error in that line is not obvious to you, I have to ask how long you've been programming. I'm not trying to be rude, but seriously, it's simply a case of unbalanced parenthesis. Change the line to this:
[php]if (!isset($_GET['page'])) {[/php]
Sorry I posted code that was not error-free, though. :)

[Narayan15]

i need help in php
<?php
$num=2;
?>
Text to be displayed

Here i m not getting $num value when i print echo $_GET['id'];

[Narayan15]

Please help me

[Troy]

i need help in php <?php $num=2; ?> Text to be displayed

Here i m not getting $num value when i print echo $_GET['id'];

Here is a tiny bit of PHP that will better demonstrate the logic. Instead of a link, I used a textbox in a form so you could actually change the ID to see that the code works. Create a page namedtest.php like this:

<?php
$id = (isset($_GET['id']) ? $_GET['id'] : 2);
?>

ID: <?= $id ?>

<form method="GET" action="test.php">
<input type="text" name="id" size="3" value="<?= $id ?>" />
<input type="submit" />

For a short time, you can see this code in action here: http://www.troywolf.com/tmp/test.php

[rekhasuresh]

sample.php

<?php
header('location:login.php?id='.$username.'');
?>

login.php

<?php
$user=htmlspecialchars($_GET["id"]);

echo $user;
?>