Hi Sunitha, There is no direct solution for your scenario. But we can do it in work around process. While uploading the file, we know the file name, path and extension (say .doc). Write seperate method such that if you pass the same file details it needs to check the file extension and try to open the same file with the related object (or some other way). If it opens then close it immediately, so we had known it is correct file if not that is corrupted file. if it is corrupted you obviously gives a message " File is Corrupted". This is a work around solution, for this you need to have seperate module to be written to check for file which is correct or corrupted one. Let me know if you have any further information. Once you done with your task plz let me know if have you used any other logic for this scenario. Regards, Srinivasa Rao G Sr. Software Engineer | 
