I'm very new to sql/hive. At first, I loaded a txt file into hive using:
drop table if exists Tran_data;
create table Tran_data(tran_time string,
resort string, settled double)
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t' LINES TERMINATED BY '\n';
Load data local inpath 'C:\Users\me\Documents\transaction_data.txt' into table Tran_Data;
The variable tran_time
in the txt file is like this:10-APR-2014 15:01. After loading this Tran_data table, I tried to convert tran_time
to a "standard" format so that I can join this table to another table using tran_time
as the join
key. The date format desired is 'yyyymmdd'. I searched online resources, and found this: unix_timestamp(substr(tran_time,1,11),'dd-MMM-yyyy')
So essentially, I'm doing this: unix_timestamp('10-APR-2014','dd-MMM-yyyy')
. However, the output is "NULL".
So my question is: how to convert the date format to a "standard" format, and then further convert it to 'yyyymmdd' format?
unix_timestamp('2014-05-01','dd-mmm-yyyy') will work, your input string should be in this format for hive yyyy-mm-dd or yyyy-mm-dd hh:mm:ss
Where as you are trying with '01-MAY-2014' hive won't understand it as a date string
unix_timestamp function will convert given string date format to unix timestamp in seconds , but not like this format dd-mm-yyyy.
You need to write your own custom udf to convert a given string date to the format that you need as present Hive do not have any predefined functions. We have to_date function to convert a timestamp to date , remaining all unix_timestamp functions won't help your problem.
input format: mmddyyyy
01032018
output after query: yyyymmdd
20180103
To help someone in the future:
The following function should work as it worked in my case
My current Hive Version: Hive 0.12.0-cdh5.1.5
I converted datetime in first column to date in second column using the below hive date functions. Hope this helps!