$_GET Question

[smithmr8]

Hello Again,

I am trying to get this script to show a specific person's details, ie Username, when their ID is put at the end of a URL. E.g.

Lets say someone with ID:3 has the username "Fred".

 

When the page, view.php?id=3 is shown, it should echo that username, 'Fred'.

 

I've tried to do it, but cant seen to remember how to use the id used in the url and assign it to a variable.

This is what I've got so far, which isn't working.

 

<?php include('header.php') ?>
<?php
if($_GET['id'] == $id){
$information="SELECT * FROM users WHERE ID = $id";
$result_information = mysql_query($information);
$view=mysql_fetch_array($result_information);

echo $view['username'];
} else {
echo "Error";
echo $id;
}


?>
<?php include('footer.php') ?>

[Lumio]

what about

<?php
if (isset($_GET['id'] && is_numeric($_GET['id'])) {
//code
}

[smithmr8]

But, how do I assign that numeric ID to a variable ?

[wildteen88]

simpley add $id = $_GET['id']; in between the parenthesis for Lumios code, eg:

 

if (isset($_GET['id'] && is_numeric($_GET['id'])) {
    $id = $_GET['id'];
}
else
{
    die('Invalid id'); // display error if id is not numeric;
}

[smithmr8]

Oh, right. I've tried that, but it doesn't seem to be working. I cant give any error details as my host doesn't have them enabled for some reason. I now have this..

if (isset($_GET['id'] && is_numeric($_GET['id'])) {
  $id = $_GET['id'];
$information="SELECT * FROM users WHERE ID = $id";
$result_information = mysql_query($information);
$view=mysql_fetch_array($result_information);

echo $view['username'];
} else {
echo "Error";
echo $id;
}

[tmyonline]

Did you establish a connection to the database ?

[smithmr8]

Its always connected to the database. I know the connections are working, as the rest of the site work fine. Its something to do with the

(isset($_GET['id'] && is_numeric($_GET['id']))

.

[Stooney]

try this

 

<?php 
if (isset($_GET['id']) && is_numeric($_GET['id'])) {
$id = $_GET['id'];
$information="SELECT * FROM users WHERE ID = $id";
$result_information = mysql_query($information);
$view=mysql_fetch_assoc($result_information);
echo $view['username'];
} else {
echo "Error";
echo $id;
}
?>

[smithmr8]

Yay. That worked. Thanks alot. . Much Appreciated.