hive和presto的求数组长度函数区别及注意事项

1、任务

获取邮箱字符串’@'后字符串 ，求长度

2、hive & spark-sql 求数组长度的函数 size

hive & spark-sql 求数组长度的函数 size

select size(split(email, '@')),split(email, '@'),split(email, '@')[0],split(email, '@')[1]
FROM 
(select "[email protected]" as email union select "[email protected]" as email) tb_mid;

select size(split(email, '@')),split(email, '@'),split(email, '@')[0],split(email, '@')[1]
FROM 
(select '[email protected]' as email union select '[email protected]' as email) tb_mid;

2    ["tom","126.com.cn"]    tom    126.com.cn
2    ["jack","126.com"]    jack    126.com
Time taken: 0.723 seconds, Fetched 2 row(s)

3、presto 求数组长度的函数 cardinality

presto  求数组长度的函数 cardinality

select cardinality(split(email, '@')),split(email, '@'),split(email, '@')[1],split(email, '@')[2]
FROM 
(select '[email protected]' as email union select '[email protected]' as email) tb_mid;

_col0 |       _col1       | _col2 |   _col3    
-------+-------------------+-------+------------
     2 | [tom, 126.com.cn] | tom   | 126.com.cn 
     2 | [jack, 126.com]   | jack  | 126.com    
(2 rows)

select cardinality(split(email, '@')),split(email, '@'),split(email, '@')[1],split(email, '@')[2]
FROM 
(select "[email protected]" as email union select "[email protected]" as email) tb_mid;

Query 20231019_070945_20009_n9u2s failed: line 3:9: Column '[email protected]' cannot be resolved
select cardinality(split(email, '@')),split(email, '@'),split(email, '@')[1],split(email, '@')[2]
FROM
(select "[email protected]" as email union select "[email protected]" as email) tb_mid

4、注意事项

1）、在计算数组长度的时候，hive和presto的函数不同
其中hive的size函数默认数组的下标从0开始
presto的cardinality函数默认数组的下标从1开始

2）、presto 不支持双引号 ，而hive 既支持单引号，也支持双引号

presto> SELECT 
     -> email,
     -> (case when cardinality(split(email, '@')) = 2 then split(email, '@')[1] else '' end ) as email_suffix
     -> FROM 
     -> (select "[email protected]" as email union select "[email protected]" as email) tb_mid;
Query 20231016_070153_17958_p9f2s failed: line 5:9: Column '[email protected]' cannot be resolved
SELECT
email,
(case when cardinality(split(email, '@')) = 2 then split(email, '@')[1] else '' end ) as email_suffix
FROM
(select "[email protected]" as email union select "[email protected]" as email) tb_mid