我已经使用hive在hbase中创建了一个表:
hive> CREATE TABLE hbase_table_emp(id int, name string, role string) STORED BY 'org.apache.hadoop.hive.hbase.HBaseStorageHandler' WITH SERDEPROPERTIES ("hbase.columns.mapping" = ":key,cf1:name,cf1:role") TBLPROPERTIES ("hbase.table.name" = "emp");
并创建了另一个表来加载数据:
hive> create table testemp(id int, name string, role string) row format delimited fields terminated by '\t'; hive> load data local inpath '/home/user/sample.txt' into table testemp;
最后将数据插入到hbase表中:
hive> insert overwrite table hbase_table_emp select * from testemp; hive> select * from hbase_table_emp; OK 123 Ram TeamLead 456 Silva Member 789 Krishna Member time taken: 0.160 seconds, Fetched: 3 row(s)
该表在hbase中如下所示:
hbase(main):002:0> scan 'emp' ROW COLUMN+CELL 123 column=cf1:name, timestamp=1422540225254, value=Ram 123 column=cf1:role, timestamp=1422540225254, value=TeamLead 456 column=cf1:name, timestamp=1422540225254, value=Silva 456 column=cf1:role, timestamp=1422540225254, value=Member 789 column=cf1:name, timestamp=1422540225254, value=Krishna 789 column=cf1:role, timestamp=1422540225254, value=Member 3 row(s) in 2.1230 seconds
我可以对JSON文件做同样的事情:
{"id": 123, "name": "Ram", "role":"TeamLead"} {"id": 456, "name": "Silva", "role":"Member"} {"id": 789, "name": "Krishna", "role":"Member"}
并做:
hive> load data local inpath '/home/user/sample.json' into table testemp;
请帮忙 !:)
您可以使用该get_json_object函数将数据解析为JSON对象。例如,如果您使用JSON数据创建登台表:
get_json_object
DROP TABLE IF EXISTS staging; CREATE TABLE staging (json STRING); LOAD DATA LOCAL INPATH '/local/path/to/jsonfile' INTO TABLE staging;
然后使用get_json_object提取要加载到表中的属性:
INSERT OVERWRITE TABLE hbase_table_emp SELECT get_json_object(json, "$.id") AS id, get_json_object(json, "$.name") AS name, get_json_object(json, "$.role") AS role FROM staging;
有此功能的更全面的讨论在这里。