我在Hive中有一个表,其中包含3个表格,如下所示;
timestamp UserID OtherId
2016-09-01 123 "101","222","321","987","393.1","090","467","863"
2016-09-01 124 "188","389","673","972","193","100","143","210"
2016-09-01 125 "888","120","482","594","393.2"
2016-09-01 126 "441","501","322","671","008","899"
2016-09-01 127 "004","700","393.4","761","467","356","643","578"
2016-09-01 128 "322","582","348"
2016-09-01 129 "029","393.8","126","187"
其中OtherID是数组。
我需要解析OtherID,以便得到的数据集如下,因为我只对包含'393%'的值感兴趣
timestamp UserID OtherId
2016-09-01 123 393.1
2016-09-01 125 393.2
2016-09-01 127 393.4
2016-09-01 129 393.8
我已经研究了大量的解析函数,但似乎它们都是为了返回值的位置,或者你需要指定值的位置来返回它。这两个选项在这里都不起作用,因为对于任何给定的行,数组中的任何点都可以出现“3309%”。 还有一个事实是我需要加入通配符以允许我想要的值的变化。
另一种选择是爆炸,但我的桌子对于那个选项来说太大了。
我认为UDF可能是唯一可行的方法,但欢迎那里有一些指导。
感谢任何帮助。
答案 0 :(得分:0)
使用hive中提供的横向视图选项,您可以轻松完成所需操作。
0: jdbc:hive2://quickstart:10000/default> select * from test_5;
+-----------+------------+----------------------------------------------+
| test_5.t | test_5.id | test_5.oid |
+-----------+------------+----------------------------------------------+
| 123 | 123 | "222","321","987","393.1","090","467","863" |
+-----------+------------+----------------------------------------------+
这就是诀窍:
SELECT id, ooid
FROM test_5
LATERAL VIEW EXPLODE(SPLIT(oid,",")) temp AS ooid;
+------+----------+
| id | ooid |
+------+----------+
| 123 | "222" |
| 123 | "321" |
| 123 | "987" |
| 123 | "393.1" |
| 123 | "090" |
| 123 | "467" |
| 123 | "863" |
+------+----------+
埃尔戈:
SELECT id, regexp_replace(ooid,'"','')
FROM test_5
LATERAL VIEW EXPLODE(SPLIT(oid,",")) temp AS ooid;
WHERE ooid LIKE '"393%';
+------+----------+
| id | ooid |
+------+----------+
| 123 | 393.1 |
+------+----------+
答案 1 :(得分:0)
可以尝试如下:
hive> select timestamp1, userid, otherids from userdet1 LATERAL VIEW explode(otherid) testTable as otherids where otherids LIKE concat('393','%');
好的
2016-09-01 123 393.1
2016-09-01 125 393.2
2016-09-01 127 393.4
2016-09-01 129 393.8
Time taken: 0.297 seconds, Fetched: 4 row(s)