PHP: If internet explorer 6, 7, 8 , or 9

Question

I want to do a conditional in PHP for the different versions of Internet Explorer along the lines of:

if($browser == ie6){ //do this} elseif($browser == ie7) { //dothis } elseif...

I have seen many variations on similar code, but looking for something super simple that is very easy to code to do some simple if and else and do different things.

Thanks

EDIT: I need this to show some different messages to users so CSS conditionals etc are no good.

Answer 1

This is what I ended up using a variation of, which checks for IE8 and below:

if (preg_match('/MSIE\s(?P<v>\d+)/i', @$_SERVER['HTTP_USER_AGENT'], $B) && $B['v'] <= 8) {     // Browsers IE 8 and below } else {     // All other browsers } 

Answer 2

A version that will continue to work with both IE10 and IE11:

preg_match('/MSIE (.*?);/', $_SERVER['HTTP_USER_AGENT'], $matches); if(count($matches)<2){   preg_match('/Trident\/\d{1,2}.\d{1,2}; rv:([0-9]*)/', $_SERVER['HTTP_USER_AGENT'], $matches); }  if (count($matches)>1){   //Then we're using IE   $version = $matches[1];    switch(true){     case ($version<=8):       //IE 8 or under!       break;      case ($version==9 || $version==10):       //IE9 & IE10!       break;      case ($version==11):       //Version 11!       break;      default:       //You get the idea   } }