table with different partitions in athena aws










0














Is it possible to create a table with different types of partitions in athena?



for example having a partition per year month day and another partition only by id



 CREATE EXTERNAL TABLE IF NOT EXISTS table_example(
name string,
adress
PARTITIONED BY (year string, month string, day string) ----> partition 1
PARTITIONED BY (id int) -----------> partition 2

ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
WITH SERDEPROPERTIES ('ignore.malformed.json' = 'true')
LOCATION 's3://example/folder/';


is something similar to this possible?



thanks










share|improve this question




























    0














    Is it possible to create a table with different types of partitions in athena?



    for example having a partition per year month day and another partition only by id



     CREATE EXTERNAL TABLE IF NOT EXISTS table_example(
    name string,
    adress
    PARTITIONED BY (year string, month string, day string) ----> partition 1
    PARTITIONED BY (id int) -----------> partition 2

    ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
    WITH SERDEPROPERTIES ('ignore.malformed.json' = 'true')
    LOCATION 's3://example/folder/';


    is something similar to this possible?



    thanks










    share|improve this question


























      0












      0








      0







      Is it possible to create a table with different types of partitions in athena?



      for example having a partition per year month day and another partition only by id



       CREATE EXTERNAL TABLE IF NOT EXISTS table_example(
      name string,
      adress
      PARTITIONED BY (year string, month string, day string) ----> partition 1
      PARTITIONED BY (id int) -----------> partition 2

      ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
      WITH SERDEPROPERTIES ('ignore.malformed.json' = 'true')
      LOCATION 's3://example/folder/';


      is something similar to this possible?



      thanks










      share|improve this question















      Is it possible to create a table with different types of partitions in athena?



      for example having a partition per year month day and another partition only by id



       CREATE EXTERNAL TABLE IF NOT EXISTS table_example(
      name string,
      adress
      PARTITIONED BY (year string, month string, day string) ----> partition 1
      PARTITIONED BY (id int) -----------> partition 2

      ROW FORMAT SERDE 'org.openx.data.jsonserde.JsonSerDe'
      WITH SERDEPROPERTIES ('ignore.malformed.json' = 'true')
      LOCATION 's3://example/folder/';


      is something similar to this possible?



      thanks







      amazon-s3 hive partition amazon-athena prestodb






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      edited Nov 12 '18 at 20:13









      Piotr Findeisen

      5,0111538




      5,0111538










      asked Nov 12 '18 at 18:19









      Bar

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          No, it is not possible.
          Partitioning is not "indexing". It's layout of the data on the file system (or file-system-like storage, here: s3).
          What you're after would be two independent copies of the data. For this, you can simply create two tables, one with partitioning by year/month/day and second with partitioning by id.



          However, assuming that id is an identifier in your table, you really do not want to partition by id. You might be interested in exploring bucketing by id, though.






          share|improve this answer




















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            1 Answer
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            active

            oldest

            votes









            1














            No, it is not possible.
            Partitioning is not "indexing". It's layout of the data on the file system (or file-system-like storage, here: s3).
            What you're after would be two independent copies of the data. For this, you can simply create two tables, one with partitioning by year/month/day and second with partitioning by id.



            However, assuming that id is an identifier in your table, you really do not want to partition by id. You might be interested in exploring bucketing by id, though.






            share|improve this answer

























              1














              No, it is not possible.
              Partitioning is not "indexing". It's layout of the data on the file system (or file-system-like storage, here: s3).
              What you're after would be two independent copies of the data. For this, you can simply create two tables, one with partitioning by year/month/day and second with partitioning by id.



              However, assuming that id is an identifier in your table, you really do not want to partition by id. You might be interested in exploring bucketing by id, though.






              share|improve this answer























                1












                1








                1






                No, it is not possible.
                Partitioning is not "indexing". It's layout of the data on the file system (or file-system-like storage, here: s3).
                What you're after would be two independent copies of the data. For this, you can simply create two tables, one with partitioning by year/month/day and second with partitioning by id.



                However, assuming that id is an identifier in your table, you really do not want to partition by id. You might be interested in exploring bucketing by id, though.






                share|improve this answer












                No, it is not possible.
                Partitioning is not "indexing". It's layout of the data on the file system (or file-system-like storage, here: s3).
                What you're after would be two independent copies of the data. For this, you can simply create two tables, one with partitioning by year/month/day and second with partitioning by id.



                However, assuming that id is an identifier in your table, you really do not want to partition by id. You might be interested in exploring bucketing by id, though.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 '18 at 20:16









                Piotr Findeisen

                5,0111538




                5,0111538



























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