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From Michele Bertoni <michele1.bert...@mail.polimi.it>
Subject Re: open multiple file from list of uri
Date Fri, 26 Jun 2015 10:25:28 GMT
Got it!
i will try thanks! :)

What about writing a section of it in the programming guide?
I found a couple of topic about the readers in the mailing list, it seems it may be helpful

Il giorno 26/giu/2015, alle ore 12:21, Stephan Ewen <sewen@apache.org<mailto:sewen@apache.org>>
ha scritto:

Sure, just override the "createInputSplits()" method. Call for each of your file paths "super.createInputSplits()"
and then combine the results into one array that you return.

That should do it...

On Fri, Jun 26, 2015 at 12:19 PM, Michele Bertoni <michele1.bertoni@mail.polimi.it<mailto:michele1.bertoni@mail.polimi.it>>
Hi Stephan, thanks for answering,
right now I am using an extension of the DelimitedInputFormat, is there a way to merge it
with the option 2?

Il giorno 26/giu/2015, alle ore 12:17, Stephan Ewen <sewen@apache.org<mailto:sewen@apache.org>>
ha scritto:

There are two ways you can realize that:

1) Create multiple sources and union them. This is easy, but probably a bit less efficient.

2) Override the FileInputFormat's createInputSplits method to take a union of the paths to
create a list of all files and fils splits that will be read.


On Fri, Jun 26, 2015 at 12:12 PM, Michele Bertoni <michele1.bertoni@mail.polimi.it<mailto:michele1.bertoni@mail.polimi.it>>
Hi everybody,
is there a way to specify a list of URI (“hdfs://file1”,”hdfs://file2”,…) and open
them as different files?
I know i may open the entire directory, but i want to be able to select a subset of files
in the directory


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