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View Full Version : session inside links or database?



textahead
03-08-2009, 10:47 AM
Heres a curly one for yous.
I have about 20 links on a page that I need to have a variable associated with it
Ie: if you click on the link <a href=USA.php>NEW YORK</A> it will take you to USA.php but display the city of New York at the top of the page. the next link would be <a href=USA.php>Detroit</A> this would also take you to the USA.php
but would display Detroit at the top of the page and so on
The way I imagine this would work is by something like this

<a href=USA.php<?php[$USA=New York]?> New York </a>
<a href=USA.php<?php[$USA=Detroit]?> Detroit</a>

i would then have to pass the variable $USA to a session so that it carrys to the next page and is displayed.

Am I on the right track or is there a better way of doing this?

I would also like other things to change not just the name at the top but If i can figure out how to carry a different value for each of 20 or so links/variable across to one page I think I could work the rest out.
Cheers in advance

snowysweb
03-08-2009, 11:01 AM
<a href=USA.php?city=detroit></a>
<a href=USA.php?city=newyork></a>

if(isset($_GET['city']))
{
if($_GET['city']=="detroit")
{
echo "Detroit";
}
elseif($_GET['city']=="newyork")
{
echo "New York";
}
}


Something like that im guessing?

snowysweb
03-08-2009, 11:09 AM
also i would lookup the switch function @ php.net instead of else if you are using more than 2.

snowysweb
03-08-2009, 11:13 AM
or create a database where all the info for the citys are in one table and pull them out via there id.



<a href="USA.php?city=1">Detroit</a>
<a href="USA.php?city=2">New York</a>


$city = $_GET['city'];
$thing = mysql_query("SELECT * from `usatable` where id='$city'");
if(mysql_num_rows($thing)!=0)
{

$thing_array = mysql_fetch_array($thing);
$city = $thing_array["city"];
$population = $thing_array["population"];
$weather = $thing_array["weather"];

}
else
{
die("No data");
}




<h1><?php echo $city; ?></h1>

and so on..



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