php - Generating AJAX callback- table on a javascript popup -

i have used jquery ajax initiates on click of button , , on click of variable passes php script jquery post using. when try append return data on javascript alert() method returns php script's html contents instead rather rendering out. can assit me on this?

<?php  $var = $_post['var'];  $sql = mysql_query("select * racers style = '$var'");   while ($r = mysql_fetch_assoc($sql))     {     $name = $r['rname'];      echo '<tr><td>'.$name.'</td></tr>';     } ?> 

you have syntax error, try this;

<?php     $var = $_post['var'];     $sql = mysql_query("select * racers style = '$var'");   while ($r = mysql_fetch_assoc($sql))  {    $name = $r['rname'];     echo '<tr><td>'.$name.'</td></tr>';  } ?> 

change sql_query mysql_query