javascript ajax file upload

jquery form url : https://github.com/malsup/form

 

<form id="uploadForm" method="post" action="/upload" enctype="multipart/form-data">        

    <input id="profile" type="file" name="file" accept="image/png, image/jpg, image/gif, image/jpeg" />    

</form>

  

<script src="${jsPath}/jquery.form.js"></script>

<script>

  function fnUpload() {   

    if(fnUpload.status == 'ing') {

      alert("처리중입니다. 잠시만 기다려주세요.");

      return false;

    } else {

      fnUpload.status = 'ing';

    }

    $("#uploadForm").ajaxForm({

      beforeSubmit : function(data, frm, opt) {

      },

      success : function(result, status) {

        if (result.code == 'SUCC') {

          alert("업로드에 성공하였습니다.");          

        } else if (result.code == 'FAIL') {

          fnUpload.status = 'fail';

          alert(result.msg);

        } else {

          fnUpload.status = 'fail';

          alert("업로드에 실패하였습니다. 잠시 후 다시 시도해주세요.");

        }

      }, 

      fail : function(){

        fnUpload.status = 'error';

      }

    }).submit();

  }

</script>  

 

 

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