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)
这个表在hbase中是这样的:
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
我可以为JSON文件执行相同的操作:
{"id": 123, "name": "Ram", "role":"TeamLead"}
{"id": 456, "name": "Silva", "role":"Member"}
{"id": 789, "name": "Krishna", "role":"Member"}
并且做:
hive> load data local inpath '/home/user/sample.json' into table testemp;
请帮忙! :)
答案 0 :(得分:2)
您可以使用get_json_object
函数将数据解析为JSON对象。例如,如果使用JSON数据创建临时表:
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;
对此功能进行了更全面的讨论here。