How about: FileSplit fileSplit = (FileSplit) context.getInputSplit(); String sFileName = fileSplit.getPath().getName();
On Sun, Jun 27, 2010 at 7:59 AM, zhangguoping zhangguoping < [email protected]> wrote: > Hi, > > how to get the current input file in a Mapper in 0.20.2 ? > > Thanks, >
