hive復合數據類型之map

概述

MAP：MAP包含key->value鍵值對，可以通過key來訪問元素。比如”userlist”是一個map類型，其中username是key，password是value；那么我們可以通過userlist['username']來得到這個用戶對應的password；

操作實例

1、創建表

hive> create table map_test(id string,perf map<string,int>)

> row format delimited fields terminated by "\t"

> collection items terminated by ','

> map keys terminated by ':'

> ;

與struct的區別：

create table map_test(id string,perf map<string,int>)
 

create table student_test(id int,info struct<name:string,age:int>)

------------

再建一張map_test2，測試map字段名不一致會如何

hive> create table map_test2(id string,perf map<string,int>)
> row format delimited fields terminated by "\t"
> collection items terminated by ','
> map keys terminated by ':'
> ;

2、準備文件

---------字段名一致-------

[root@hello110 data]# vi ?map_test

1001 ? ?job:80,team:123,person:700
1002 ? ?job:90,team:234,person:800
1003 ? ?job:70,team:345,person:900
1004 ? ?job:60,team:456,person:1000
1005 ? ?job:59,team:678,person:844
1006 ? ?job:98,team:832,person:866

---------字段名不一致------

[root@hello110 data]# vi ?map_test2
1001 ? ?job_1:80,team:123,person:700
1002 ? ?job_2:90,team:234,person:800
1003 ? ?job:70,team:345,person:900
1004 ? ?job:60,team:456,person:1000
1005 ? ?job:59,team:678,person:844
1006 ? ?job:98,team:832,person:866

3、文件導入表

load data local inpath "/data/map_test" into table map_test;

load data local inpath "/data/map_test2" into table map_test2;

4、查看表

------map_test表-----

hive> select * from map_test;

OK

1001 {"job":80,"team":123,"person":700}

1002 {"job":90,"team":234,"person":800}

1003 {"job":70,"team":345,"person":900}

1004 {"job":60,"team":456,"person":1000}

1005 {"job":59,"team":678,"person":844}

1006 {"job":98,"team":832,"person":866}

Time taken: 2.264 seconds, Fetched: 6 row(s)

---------用[''] 的方式獲取-----------

hive> select perf['job'],perf['team'],perf['person'] from map_test;

OK

80 123 700

90 234 800

70 345 900

60 456 1000

59 678 844

98 832 866

Time taken: 4.377 seconds, Fetched: 6 row(s)

hive> select perf['job'],perf['team'],perf['person123'] from map_test;

OK

80 123 NULL

90 234 NULL

70 345 NULL

60 456 NULL

59 678 NULL

98 832 NULL

Time taken: 4.212 seconds, Fetched: 6 row(s)

hive> select perf from map_test;

OK

{"job":80,"team":123,"person":700}

{"job":90,"team":234,"person":800}

{"job":70,"team":345,"person":900}

{"job":60,"team":456,"person":1000}

{"job":59,"team":678,"person":844}

{"job":98,"team":832,"person":866}

Time taken: 4.118 seconds, Fetched: 6 row(s)

hive> 

----------map_test2-------

hive (default)> select * from map_test2;
OK
map_test2.id ? ?map_test2.perf
1001 ? ?job_1:80,team:123,person:700 ? ?NULL
1002 ? ?job_2:90,team:234,person:800 ? ?NULL
1003 ? ?job:70,team:345,person:900 ? ? ?NULL
1004 ? ?job:60,team:456,person:1000 ? ? NULL
1005 ? ?job:59,team:678,person:844 ? ? ?NULL
1006 ? ?job:98,team:832,person:866 ? ? ?NULL
Time taken: 0.086 seconds, Fetched: 6 row(s)

----------------
hive (default)> select perf['job_1'] from map_test2;
OK
c0
NULL
NULL
NULL
NULL
NULL
NULL
Time taken: 0.284 seconds, Fetched: 6 row(s)

-----------------------
hive (default)> select perf['job'] from map_test2;
OK
c0
NULL
NULL
NULL
NULL
NULL
NULL
Time taken: 0.084 seconds, Fetched: 6 row(s)

-------------------------------
hive (default)> select perf['job_2'] from map_test2;
OK
c0
NULL
NULL
NULL
NULL
NULL
NULL
Time taken: 0.074 seconds, Fetched: 6 row(s)

------------------------------
hive (default)> select perf from map_test2;
OK
perf
NULL
NULL
NULL
NULL
NULL
NULL
Time taken: 0.076 seconds, Fetched: 6 row(s)

-----------------原來：如果map字段名不一致，會被當成id字段來處理了---------
hive (default)> select id from map_test2;
OK
id
1001 ? ?job_1:80,team:123,person:700
1002 ? ?job_2:90,team:234,person:800
1003 ? ?job:70,team:345,person:900
1004 ? ?job:60,team:456,person:1000
1005 ? ?job:59,team:678,person:844
1006 ? ?job:98,team:832,person:866
Time taken: 0.064 seconds, Fetched: 6 row(s)

5、hadoop中的文件內容

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