Display image depending on value of a variable

Question

Hi,

I cannot work out how to display an image depending on the vaule of a product($idd):

    foreach ($response['products'] as $product) 
{ 
 if($filter != "all" && $bb[$product['product-id']['value']] && $bb[$product['product-id']['value']][0] != $filter) continue; 

 $r = array(); $idd = 0; $ca = 0; 

  foreach($product as $key=>$value) 
 { 
  if($key == "product-id") 
  { 
   $idd = (int)$value['value']; 
   $r[0] = $idd?"<input type='radio' name='product_id' value='" . $idd . "'/>":"0&nbsp;"; 
   //price// 
    $r[4] = $bb[$idd]?$bb[$idd][3]:"ID"; 
   //info// 
    $r[5] = "<a href='checker/info.php?id=" .$idd. "'; rel=\"shadowbox;width=400;height=200\" class=\"option\"'><img src=\"http://www.flipconnect.co.uk/images/more-info.gif\" style=\"border:none; padding-top:24px;\" alt=\"more information about this pack\" /></a>"; 
   //buy 
    $r[6] = "<a href='checker/next.php?id=" . $idd . "&amp;postcode=".$postcode."&amp;speed=".$phone."'><img src=\"http://www.flipconnect.co.uk/images/buy-now.gif\" style=\"border:none\" alt=\"buy this package\" /></a>"; 
  }

I'm thinking it should tak the form of:

$r[0] = if ($idd == 1021) echo "IMAGE URL": echo "";

But doesn't work, by the way all the $idd are stored in an array with a load of othewr info.

Answer 1

Your question is not quite clear to me but Yours this line of code

`$r[0] = if ($idd == 1021) echo "IMAGE URL": echo "";`

can be rewritten as

echo $r[0] = ($idd == 1021) ? "IMAGE URL" : "";

Try it.

EDIT

Regarding Bifter's comment "This work, how would adapt to display a different image if the $idd value is different. i.e - $idd == 1021 - Image 1 is displayed $idd == 1022 - Image 2 is displayed $idd == 1023 - Image 3 is displayed"

if your data array is something like ..

$arr = array('1021' => 'img 1', '1022' => 'Img 2')

then you could try something like this

echo $r[0] = array_key_exists($idd, $data) ? $data[$id] : "";

assuming that you are looping through ID values ($id).