laravel框架之即时更改

Posted 2020-12-02 azzy

表单//
@foreach($res as $k=>$v)
 <tr    id="{{$v->id}}" >
     <td>{{$v->id}}</td>
     <td >{{$v->tye}}</td>
     <td   fd="name" > {{$v->name}}</td>
     <td><a href="#"    class="shan"  id="{{$v->id}}"     >删除</a></td>
 </tr>
//jqery

<script>
   $(document).on("dblclick","td",function () {
         fd=$(this).attr("fd");
         id=$(this).parents().attr("id");
          text=$(this).text();
            that=$(this);
            if (fd==undefined){
                alert("不能修改")
            }
         html=$(this).html(‘<input type="text"   class="gg"  value="‘+text+‘">‘)
        $(".gg").blur(function () {
            new_val=$(this).val();
             $.ajax({
                 url:"zt",
                 dataType:"json",
                 type:"get",
                 data:{
                     fd:fd,
                     id:id,
                     new_val:new_val
                 },
                 success:function (data) {
                   if (data==1){
                      that.html(new_val)
                   }
                 }

             })
        }) 
   })
</script>

//控制器

function  zt(req $req){
    $fd=$req->get("fd");
    $id=$req->get("id");
    $new_val=$req->get("new_val");
   $res= DB::table("ftyp")->where("id",$id)->update([$fd=>$new_val]);
   return 1;
}