jQuery 이것도 공부해야해
$.ajax fileupload with formData

<input type="file" name="upfile" id="upfile">
<button id="up">upload </button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>
   $('#up').on('click', function(){
       let formData = new FormData();
       formData.append('title', 'file upload test with $.ajax');
       formData.append('upfile', $('#upfile')[0].files[0]);

        let data = [];
        formData.forEach(((value, key)=> data[key]= value));
        console.log(data);
       
       $.ajax({
         url: '/test/regist',
         dataType: 'json',
         enctype: 'multipart/form-data',
            type: 'post',
            processData: false, //필수
            contentType: false, //필수
            data: formData,
            success: function(res){},
            error: function(err){}
      });
   });
</script>
 

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