增量数据和合并问题验证

1.建立基表和增量测试数据

[root@node1 delta_merge]# pwd
/root/delta_merge
[root@node1 delta_merge]# cat base.txt
1001,gongshaocheng
1002,LIDACHAO
[root@node1 delta_merge]# cat delta.txt
1002,lidachao
1003,chenjianzhong

[root@node1 delta_merge]# hdfs dfs -mkdir /user/merge_delta
[root@node1 delta_merge]# hdfs dfs -mkdir /user/merge_delta/base
[root@node1 delta_merge]# hdfs dfs -mkdir /user/merge_delta/delta
[root@node1 delta_merge]# hdfs dfs -put base.txt /user/merge_delta/base
[root@node1 delta_merge]# hdfs dfs -put delta.txt /user/merge_delta/delta

2.建立测试表

hive> create external table base(id string,name string)
> ROW FORMAT DELIMITED FIELDS TERMINATED BY ','
> location "/user/merge_delta/base/";
OK
Time taken: 0.304 seconds
hive> select * from base;
OK
1001    gongshaocheng
1002    LIDACHAO
Time taken: 0.875 seconds, Fetched: 2 row(s)
hive> create external table delta(id string,name string)
> ROW FORMAT DELIMITED FIELDS TERMINATED BY ','
> location "/user/merge_delta/delta/";
OK
Time taken: 0.134 seconds
hive> select * from delta;
OK
1002    lidachao
1003    chenjianzhong
Time taken: 0.321 seconds, Fetched: 2 row(s)

3.测试：

a. full outer join语法：

hive> select base.*,delta.* from base full outer join delta on base.id = delta.id;

结果如下：

1001    gongshaocheng    NULL    NULL
1002    LIDACHAO    1002    lidachao
NULL    NULL    1003    chenjianzhong

我们最终想要的答案应该是：

1001 gongshaocheng --代表保持不变的记录

1002 lidachao --代表修改后的最新记录

1003 chenjianzhong --代表新增记录

b.coalesce函数:

select
coalesce(base.id, delta.id)
from base full outer join delta on base.id = delta.id
where (delta.id is NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NUll);

结果:

1001
1002
1003

上面验证了对于主键列，我们可以采用coalesce函数，使得结果集中主键列总是有值的

c.if函数

select
if(delta.id is NULL, base.name,delta.name)
from base full outer join delta on base.id = delta.id
where (delta.id is NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NUll);

结果：

gongshaocheng
lidachao
chenjianzhong

上面验证了对于普通列，如果是未修改的数据（delta.id is NULL）,则直接用基表里的值，否则直接用增量表的数据

最后综合起来，得到我们想要的HQL语句：

select
coalesce(base.id, delta.id),
if(delta.id is NULL, base.name,delta.name)
from base full outer join delta on base.id = delta.id
where (delta.id is NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NUll);

结果如下：

hive> select
> coalesce(base.id, delta.id),
> if(delta.id is NULL, base.name,delta.name)
> from base full outer join delta on base.id = delta.id
> where (delta.id is NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NOT NUll) OR (delta.id is NOT NULL AND base.id is NUll);
Query ID = root_20151230235050_befa6322-f78f-4166-8bbd-4fde04a1a9b1
Total jobs = 1
Launching Job 1 out of 1
Number of reduce tasks not specified. Estimated from input data size: 1
In order to change the average load for a reducer (in bytes):
set hive.exec.reducers.bytes.per.reducer=<number>
In order to limit the maximum number of reducers:
set hive.exec.reducers.max=<number>
In order to set a constant number of reducers:
set mapreduce.job.reduces=<number>
Starting Job = job_1451024710809_0005, Tracking URL = http://node1.clouderachina.com:8088/proxy/application_1451024710809_0005/ Kill Command = /opt/cloudera/parcels/CDH-5.4.7-1.cdh5.4.7.p0.3/lib/hadoop/bin/hadoop job  -kill job_1451024710809_0005
Hadoop job information for Stage-1: number of mappers: 2; number of reducers: 1
2015-12-30 23:51:04,904 Stage-1 map = 0%,  reduce = 0%
2015-12-30 23:51:13,245 Stage-1 map = 100%,  reduce = 0%, Cumulative CPU 2.69 sec
2015-12-30 23:51:22,685 Stage-1 map = 100%,  reduce = 100%, Cumulative CPU 5.16 sec
MapReduce Total cumulative CPU time: 5 seconds 160 msec
Ended Job = job_1451024710809_0005
MapReduce Jobs Launched:
Stage-Stage-1: Map: 2  Reduce: 1   Cumulative CPU: 5.16 sec   HDFS Read: 12293 HDFS Write: 52 SUCCESS
Total MapReduce CPU Time Spent: 5 seconds 160 msec
OK
1001    gongshaocheng
1002    lidachao
1003    chenjianzhong
Time taken: 31.376 seconds, Fetched: 3 row(s)

注意：上面所有的HQL都只需要有一个MR作业。这就是本解决方案的精髓所在！

最后对HQL进行进一步优化：之前为了保持逻辑上的清晰，增加了WHERE子句，对FULL OUTER JOIN的三种情况进行分布讨论，但实际上两个OR合并后就是全集，其实WHERE子句是多余的。最终的HQL为:

select
coalesce(base.id, delta.id),
if(delta.id is NULL, base.name,delta.name)
from base full outer join delta on base.id = delta.id;