php Dynamic page

Question

I'm not sure if the title here is appropriate and my explanation might be just as bad but here it goes... I am using the following code to generate web pages:
source: Dynamic inclusion in PHP

<?php

$id = $_GET['id']; 

$display = $_GET['display'];
    $displays = array('page1', 'page2', 'page3', 'page4', 'page5&amp;id=$id');

if (!empty($display)) {
        if(in_array($display,$displays)) {
            $display .= '.php';
            include($display);
        }
        else {
        echo 'Page not found. Return to
        <a href="index.php">Index</a>';
        }
    }
    else { //show html

?>

a typical page:

www.website.com/dir/index.php?display=page4

My problem is this: I want to add a page to the array of allowed pages that has a dynamic value. You can see my attempt at this in the code above where i added: 'page5&id=$id'

However when i go to this page:

www.website.com/dir/index.php?display=page5&id=2

I get the error message "Page not found. Return to Index". (The table row id with value of 2 does exist in the database.)

ConcurrentHashMap · Accepted Answer · 2013-01-26 14:31:32Z

2

You should better handle the display and ID values separately. For example like this:

<?php

$display = $_GET['display'];
$displays = array('page1', 'page2', 'page3', 'page4', 'page5');

if (!empty($display)) {
        if(in_array($display,$displays)) {
            $display .= '.php';
            include($display);
        }
        else {
        echo 'Page not found. Return to
        <a href="index.php">Index</a>';
        }
    }
    else { //show html

?>

and in display5.php:

 <?php

 // some other initializations

 $id = $_GET['id'];
 if($id == 2) { // make some special actions for id = 2

 // show more html
 ?>

answered Jan 26, 2013 at 14:31

ConcurrentHashMap

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4 Comments

Hezerac Over a year ago

But if the value of id is not passed in the url how can we use $_GET['id']; ? also the value "2" I used as just an arbitrary example to represent a value determined by user navigation.

ConcurrentHashMap Over a year ago

If your request goes to www.website.com/dir/index.php?display=page5&id=2, as you wrote, the two available $_GET variables will be $_GET['display'] and $_GET['id']. So the ID will be passed inside your snippet. Every file, that gets included there, can also access the $_GET variables, so the display5.php can access it. The if($id == 2) { was just an example of me for showing off how to deal with the ID. You could do some database SELECT statements here instead, e.g.

Hezerac Over a year ago

Ok I understand. But now, for whatever reason when I try what you explained it doesnt work even though with the url as in the example i cant even $id = $_GET['id']; echo $id;

Hezerac Over a year ago

Ack. There was a typo. It works now thank you very much for the help!

phant0m · Accepted Answer · 2013-01-26 14:24:26Z

1

Just like you will receive a value in $_GET['display'], you will receive another in $_GET['id'] with a value of 2!

PHP separates them from the query string. When you use in_array(), you'll check whether page5 is in $displays, as you can see from your code, it's not.

I suggest you use var_dump($_GET); and then look at the source of the produced HTML to see how GET parameters are being handled.

answered Jan 26, 2013 at 14:24

phant0m

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Comments

take · Accepted Answer · 2013-01-26 14:24:46Z

1

Thats why in $_GET["display"] stands only "page5" and in $_GET["id"] the 2. You can check the ID in the page5.php.

For example in page5:

<? if (empty($_GET["id"]) || !is_numeric($_GET["id"])) { die("ID isn't a number!"); } ?>

answered Jan 26, 2013 at 14:24

take

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