Display Image based on URL GET

Question

Im sure this is simple but newbie here.

I have 5 URLS

cards.php&type=one
cards.php&type=two
cards.php&type=three
cards.php&type=four

Question.

How will I display an image based on the URL

Image Directory = images/cards/one.jpg,two.jpg etc...

Thanks your help is much appreciated

Answer 1

Try this:

echo "<img src='{$_GET['type']}.jpg' />";

And with a check:

if (isset($_GET['type']) && (trim($_GET['type']) != "")) {
   echo "<img src='". $_GET['type'] .".jpg";
}

Answer 2

$type = $_REQUEST['type'];
    if($type == "one")
    {
      $imag = "images/cards/one.jpg";
    }
    else if($type == "two")
    {
      $imag = "images/cards/two.jpg";
    }......

or use

switch ($type)
{
case one:
  $imag = "images/cards/one.jpg";
  break;
case two:
  $imag = "images/cards/two.jpg";
  break;
case three:
  $imag = "images/cards/three.jpg";
  break;
default:
  $imag = "images/cards/default.jpg";
}

try this one i think it might help you.

Answer 3

First of, the delimiter between the "hier-part" and the query string is ? , not & , cf. RFC 3986 - Uniform Resource Identifier (URI): Generic Syntax . Your URLs should thus look like:

cards.php?type=one
cards.php?type=two
cards.php?type=three
cards.php?type=four

Now, inspect the contents of the global $_GET array using eg print_r($_GET) . You should find an entry with the key type corresponding with the query string parameter.

Next, use the value of type ( $_GET['type'] ) to write an image tag. Beware that users can easily manipulate the query string so some input sanitation should be performed.