-Re: get name of file in mapper output directory
Luca Pireddu 2011-05-23, 12:51
The path is defined by the FileOutputFormat in use. In particular, I think
this function is responsible:
It should give you the file path before all tasks have completed and the output
is committed to the final output path.
On May 23, 2011 14:42:04 Joey Echeverria wrote:
> Hi Mark,
> FYI, I'm moving the discussion over to
> [EMAIL PROTECTED] since your question is specific to
> You can derive the output name from the TaskAttemptID which you can
> get by calling getTaskAttemptID() on the context passed to your
> cleanup() funciton. The task attempt id will look like this:
> You're interested in the m_000005 part, This gets translated into the
> output file name part-m-00005.
> On Sat, May 21, 2011 at 8:03 PM, Mark question <[EMAIL PROTECTED]> wrote:
> > Hi,
> > I'm running a job with maps only and I want by end of each map
> > (ie.Close() function) to open the file that the current map has wrote
> > using its output.collector.
> > I know "job.getWorkingDirectory()" would give me the parent path of the
> > file written, but how to get the full path or the name (ie. part-00000 or
> > part-00001).
> > Thanks,
> > Mark
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