What is the best way to write code (ajax)?

Question

JavaScript

Good afternoon, I wrote the code and everything works for me, but it looks bad, can you advise me how to make it better?

$('#input-search').keyup(function(tovar){
  let search = $(this).val();
  if(search != '')
  {
   find(search);
  }
  else
  {
   find();
  }
 function find(tovar)
 {
  $.ajax({
   url:"temp/search.php",
   method:"POST",
   data:{tovar:tovar},
   success:function(data)
   {
    $('.blocktovar').html(data);
    $(".viewtovar ").click(function(){
   let view =  $(this).attr('id');
   $.ajax ({
      url:"temp/view.php",
      method:"POST",
      data:{view:view},
      success:function (data){
          $(".blocktovar").html(data);
          
      }
   });
   });
   }
  });
 }
 });

In short, I made a search engine, and it worked fine, but the full description of the goods stopped working, so I decided to add the code that displays the product on a separate page at the end of the function that is triggered when the ajax code of the search engine is executed

$(".viewtovar ").click(function(){
   let view =  $(this).attr('id');
   $.ajax ({
      url:"temp/view.php",
      method:"POST",
      data:{view:view},
      success:function (data){
          $(".blocktovar").html(data);
          
      }
   });
   });

The output code itself is on a separate page.
I var_dump and ctrl shitv i tracked down what was the problem and this is exactly my js file