PHP에서 jQuery 작업의 일괄 삭제를 구현하는 방법

<?php 
 header(&#39;content-type:text/html;charset=utf-8&#39;);
 $pdo=new PDO(&#39;mysql:host=localhost;dbname=***;&#39;,&#39;root&#39;,&#39;root&#39;);
 $pdo->exec(&#39;set names utf8&#39;);
 $sql=&#39;select * from ***&#39;;
 $info=$pdo->query($sql)->fetchAll(PDO::FETCH_ASSOC);
?>
<center>
<table border="1">
 <tr>
  <td>id</td>
  <td>title</td>
  <td>content</td>
 </tr>
 <?php foreach($info as $k => $v){ ?>
 <tr>
  <td><input type="checkbox" name="box" value="<?= $v[&#39;id&#39;] ?>"><?= $v[&#39;id&#39;] ?></td>
  <td><?= $v[&#39;title&#39;] ?></td>
  <td><?= $v[&#39;content&#39;] ?></td>
 </tr>
 <?php } ?>
</table>
 <button>批量删除</button>
</center>
<script src="../jquery.1.12.min.js"></script>
<script>
 $(function(){
  $(&#39;button&#39;).click(function(){
   var ids=$(&#39;:checkbox&#39;);
   var str=&#39;&#39;;
   var count=0;
   for(var i=0;i<ids.length;i++){
    if(ids.eq(i).is(&#39;:checked&#39;)){
     str+=&#39;,&#39;+ids.eq(i).val();
     count++;
    }
   }
   var str=str.substr(1);
   if(confirm(&#39;你确定要删除这&#39;+count+&#39;条数据吗？&#39;)){
    //获取id后删除
    $.ajax({
     type:&#39;get&#39;,
     url:&#39;adminDel.php&#39;,
     data:{str:str},
     success:function(res){
      if(res>0){
       alert(&#39;删除成功&#39;);
       for(var i=ids.length-1;i>=0;i--){
        if(ids.eq(i).is(&#39;:checked&#39;)){
         ids.eq(i).parent().parent().remove();
        }
       }
      }
     }
    })
   }
   return false;
   /*var box=document.getElementsByName(&#39;box&#39;);
   var str="";
   for(var i=0;i<box.length;i++){
    if(box[i].checked==true){
     str+=&#39;,&#39;+box[i].value;
    }
   }
   var str=str.substr(1);
   alert(str);*/
  });
 })
</script>

<?php 
 header(&#39;content-type:text/html;charset=utf-8&#39;);
 $str=$_GET[&#39;str&#39;];
 $pdo=new PDO(&#39;mysql:host=localhost;dbname=***;&#39;,&#39;root&#39;,&#39;root&#39;);
 $pdo->exec(&#39;set names utf8&#39;);
 $sql=&#39;delete from *** where id in (&#39;.$str.&#39;)&#39;;
 $res=$pdo->exec($sql);
 //受影响行数
 echo $res;
?>