MYSQL on a page with "variabled url"

Putting my other question aside ( [webmasterworld.com...] ), because i really want to learn something from this, i proceed with this:

I want to extract contents from a mysql table. I want that based on the url that the user goes to, the content changes.

Example:
table name: test
Columns: id, name, lastname
Row: 1234, My, Name

If the user goes to page.php?id=1234 I want the page do display Name: My
Last Name: Name

So basically, i want the script to get a column (in this case the ID column) and change the contents of the page, based on the part in the row that corresponds to that column. This is all done through the URL.

Im not sure if im explaining this enough so if you need any more information, please ask. In my other thread i was asking about a specific situation on my website. This one is more general because i want to learn, and then implement that to MY situation on my own.

Thanks in advance,

Pedro S Lopez

[edited by: eelixduppy at 12:20 pm (utc) on Mar 29, 2012]
[edit reason] linked url [/edit]

$output=$fname=$lname=$null;
if (isset($_GET['id']) and is_numeric($_GET['id']) and ($_GET['id'] > 0)) {
// You can use *, but specific field names are often more efficient
// One of the rare occasions raw input is OK because we've already verified it's a number
$query = "select `name`,`lastname` from `test` where id=".$_GET['id'];
$result = mysql_query($query) or die("Cannot execute $query");
if ($row = mysql_fetch_array($result)) {
// You may want these individually - see below
$fname = $row['name'];
$lname = $row['lastname'];
$output = "$fname $lname";
}
else {
$output = 'No results were found with that query.';
}
}
else { $output = 'Invalid input supplied.'; }

MYSQL on a page with "variabled url"

psl101

rocknbil

psl101

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