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Dec 9th, 2018
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  1. var domForm = document.getElementById('ELEMENT-ID');
  2. var dataForm = new FormData(domForm);
  3.  
  4. $.ajax({
  5. type: 'post',
  6. url: URL,
  7. data: dataForm,
  8. processData: false,
  9. contentType: false,
  10. success: function (data) {
  11. console.log(data);
  12. }
  13. });
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