Sunil Sudhansu

Sunil Sudhansu

  • 1.4k
  • 307
  • 32.8k

var filename = $('#fileData').val()

Feb 6 2018 6:18 AM
var filename = $('#fileData').val();
 
it gives like this C:\fakepath\SO_Status.xlsx
 
but what i realy wanted is path of a file and not data inside a file / no need to upload a file C:\filesFolder\SO_Status.xlsx (assuming 'SO_Status.xlsx' file present in fakepath)
 
thanks

Answers (4)