Displaying a link.

[mikehillside]

Hello, this is my second day working with php, and I am trying to display a link which a user can click once their information is written to a table. The php code stores the information on the table and then returns a page with my message thanking the user, but now I need a link so they can progress through the web site.

Any information is greatly appreicated.

here is the code:

<?
$db = mysql_connect("localhost","xxxx","xxxx");

mysql_select_db (testing);

$result = mysql_query ("INSERT INTO testing_db (compname, trainname, address, city,)
VALUES ('$compname', '$trainname', '$address', '$city',)");
if(!$result)
{
echo "<b>Please try again, your information was not received:</b> ", mysql_error();
exit;
}
if($result)
{
mysql_close($db);
print "Your information was received, thank you.";
}
?>

[zalath]

Assuming I understood your agenda, you wish to insert the link if the query succeeds. In which case all you need to do is print out the appropriate HTML (which would be <a href="next.php">Proceed</a>, replace the linked file name and link text with you own).

If you use print or echo and double-quote the string, you can use \" to insert double quotes in the string. If you use single quotes, be aware that variable names will not be replaced with their values.

After that, I suggest you take a tutorial.