php - Display table data if -

im not sure if you'll i'm asking, i'll try specific , clear possible. have php code :

$result = mysql_query("select * batai svarbumas=1 order id"); while ($row = mysql_fetch_array($result)) {     extract($row);     $link=$row["linkas"];     echo "<div class='col-md-4'>";      list($width, $height) = getimagesize($link);      if($width > $height) {         echo "<img src="$link" style='height:234px;margin-left:-33%'>";     }     else if($width < $height) {         echo "<img src="$link" style='width:234px;margin-top:-33%>";     }      echo "</div>"; } 

what want do, center given image database div, example, if given images width bigger height, echos left margin. problem is, page displays 2 of elements. one, height bigger width , one, width bigger height.

if don't type ifs , echo image, every image database gets display. hope understand question , thankful advice :)

you need escape double quotes, or use single quotes , concatenate.

if($width > $height) {         echo "<img src=\"$link\" style='height:234px;margin-left:-33%'>";     }     else if($width < $height) {         echo "<img src=\"$link\" style='width:234px;margin-top:-33%'>";     } 

or

   if($width > $height) {             echo '<img src="' . $link . '" style="height:234px;margin-left:-33%">';         }         else if($width < $height) {             echo '<img src="' . $link . '" style="width:234px;margin-top:-33%">';         } 

error reporting have given message such as

parse error: parse error, expecting ','' or';'' in

so in future should turn on error reporting, http://php.net/manual/en/function.error-reporting.php.