使用ajax使用单个函数显示两个不同div中的两个表的结果

Question

$(".search").click(function() {
       a=  $(".input").val();
        $.post("search_input.php", {search:a}, function(resp) {
        if (resp!= 'incorrect') {
       $('#first').html(resp);
         }
       else{
     $('#first').html('no result found');
     }
   })

})

<input type="text" class="input">
<button class="search">Search</button>
<div id="first"></div>
<div id="second"></div>

At search_input.php, I tried searching two different tables with search.

$result1 = mysqli_query($this->con, "SELECT * from names_tb WHERE name LIKE '%$search%'");
$result2 = mysqli_query($this->con, "SELECT * from places_tb WHERE location LIKE '%$search%'");

My major issue is this, how can I display the result of result1 in the first div and the result for result2 in the second div using php? I tried using echo

$first=mysqli_num_rows($result1);
        if ($first>0) {

while ($b=$result1->fetch_array()) {

            $name=$b['name'];
            $age= $b['age'];
         }
} else{
            echo "incorrect";
        }
 $second=mysqli_num_rows($result2);
        if ($second>0) {

while ($c=$result2->fetch_array()) {

            $location=$b['location'];
            $state= $b['state'];
         }
} else{
            echo "incorrect";
        }   

But the problem with this is that, it will only display one of the results in the first div leaving the second div empty. Again, my main issue is to display result1 in the first div and result2 in the second div, how can I do this?

</div>