Hadoop 如何在不插入任何记录的情况下,通过配置单元表模式推断拼花地板模式?

Hadoop 如何在不插入任何记录的情况下,通过配置单元表模式推断拼花地板模式?,hadoop,hive,parquet,Hadoop,Hive,Parquet,现在给出一个配置单元表及其模式,即: hive> show create table nba_player; OK CREATE TABLE `nba_player`( `id` bigint, `player_id` bigint, `player_name` string, `admission_time` timestamp, `nationality` string) ROW FORMAT SERDE 'org.apache.hadoop.hive

现在给出一个配置单元表及其模式,即:

hive> show create table nba_player;
OK
CREATE TABLE `nba_player`(
  `id` bigint, 
  `player_id` bigint, 
  `player_name` string, 
  `admission_time` timestamp, 
  `nationality` string)
ROW FORMAT SERDE 
  'org.apache.hadoop.hive.ql.io.parquet.serde.ParquetHiveSerDe' 
STORED AS INPUTFORMAT 
  'org.apache.hadoop.hive.ql.io.parquet.MapredParquetInputFormat' 
OUTPUTFORMAT 
  'org.apache.hadoop.hive.ql.io.parquet.MapredParquetOutputFormat'
LOCATION
  'hdfs://endpoint:8020/user/hive/warehouse/nba_player'
TBLPROPERTIES (
  'transient_lastDdlTime'='1541140811')
Time taken: 0.022 seconds, Fetched: 16 row(s)
如何在不插入任何记录的情况下推断其拼花模式

拼花地板模式如下所示:

message_meta
{optional int64 id;
 optional int64 player_id;
 optional binary player_name;
 optional timestamp admission_time;
 optional binary nationality;}

在读取配置单元源代码时找到了解决方案@请参见org.apache.hadoop.hive.ql.io.parquet.convert.HiveSchemaConverter#convert(),您是如何使用HiveSchemaConverter的?它是否可以适用于反之亦然?基于拼花地板模式推断蜂巢模式的意义?