Display selected record that has been passed to URL

[Ren]

Hello all,

I have a page which can list all records from a table in my local database.

For each record that is displayed there is link next to it. When the link is
pressed another page opens up and the ID (automated primary key in my table)
of the record selected is passed to the URL.

So now the URL looks something like this:

http://localhost/Webpages/newpage.php?recordID=3

How do I display the details of the record that has been selected on this
new page?

Any help much appreciated!

Thanks
Ren

[jsaint]

Ren:

I'm not sure what you're looking for. Basically, your newpage.php will
use code similar to what you must have used on the first page you
describe - the one that lists the records. You will get your primary
key by looking at $_GET["recordID"] and assigning it to a variable
(like $id). Of course you will want to validate this value and make
sure it is an integer and nothing else (search for sql injection
attacks to see why).

Then, you will execute a select against your table like (SELECT * FROM
table WHERE recordID = $id) using the passed recordID. Use the
appropriate functions/syntax (these will differ if you are using PEAR
DB or native php functions, and depending on how you want to display
the data) to navigate through your result set and write out the data as
html.

HTH,
Josh

[Ren]

Thanks for the quick reply.

It sounds easy from what you have stated but I still can't seem to get it to
work.

I think it may be something to do with my select statement. Here is a
sample of the code.

$RID = $_GET['recordID'];

$query = "SELECT * FROM dvd WHERE id = '$RID'";
$result = mysql_query($qu ery);
print ("$result[title]");

Just to let you know that the table is called 'dvd' and 'id' and 'title' are
column names in my table.

I am just trying to display the 'title' of the selected record.

Thanks
Slack

[Hemanth]

Ren wrote:

Thanks for the quick reply.

It sounds easy from what you have stated but I still can't seem to get it to
work.

I think it may be something to do with my select statement. Here is a
sample of the code.

$RID = $_GET['recordID'];

$query = "SELECT * FROM dvd WHERE id = '$RID'";
$result = mysql_query($qu ery);
print ("$result[title]");

Just to let you know that the table is called 'dvd' and 'id' and 'title' are
column names in my table.

I am just trying to display the 'title' of the selected record.

Thanks
Slack

..........try this (change hostname, username, password, database_name
accordingly).

$RID = $_GET['recordID'];
$query = " SELECT * FROM dvd WHERE id = '$RID' ";

$dbcnx = @mysql_connect( "localhost" , "username", "password") ;
or die(mysql_error ());

if (!@mysql_select _db("database_n ame", $dbcnx)) die(mysql_error ());

if (!($result = @mysql_query($q uery, $dbcnx)) ) die(mysql_error ());

while($row = @mysql_fetch_ro w($result))
{
echo $row['title'];
echo "<br>";
}

if(!(mysql_clos e($dbcnx))) die(mysql_error ());
HTH,
Hemanth

[Adam Plocher]

Ren, I have written up an article on SQL Injections that you might want
to look at, too. A SQL Injection is caused when an end-user modifies
your input value and adds their own SQL to it (which could result in a
deleted DB table or worse). For this particular example you could do
something simple like

if(!is_numeric( $RID))
{
die("Invalid Input");
}

More information is here:
http://www.digitalpropulsion.org/blo...PHP_with_MySQL

[Ren]

Thanks Adam for the link.