[string] Hive cast string to date dd-MM-yyyy

How can I cast a string in the format 'dd-MM-yyyy' to a date type also in the format 'dd-MM-yyyy' in Hive?

Something along the lines of:

CAST('12-03-2010' as date 'dd-mm-yyyy')

If I have understood it correctly, you are trying to convert a String representing a given date, to another type.

Note: (As @Samson Scharfrichter has mentioned)

• the default representation of a date is ISO8601
• a date is stored in binary (not as a string)

There are a few ways to do it. And you are close to the solution. I would use the CAST (which converts to a DATE_TYPE):

SELECT cast('2018-06-05' as date); 

Result: 2018-06-05 DATE_TYPE

or (depending on your pattern)

select cast(to_date(from_unixtime(unix_timestamp('05-06-2018', 'dd-MM-yyyy'))) as date)

Result: 2018-06-05 DATE_TYPE

And if you decide to convert ISO8601 to a date type:

select cast(to_date(from_unixtime(unix_timestamp(regexp_replace('2018-06-05T08:02:59Z', 'T',' ')))) as date);

Result: 2018-06-05 DATE_TYPE

Hive has its own functions, I have written some examples for the sake of illustration of these date- and cast- functions:

Date and timestamp functions examples:

Convert String/Timestamp/Date to DATE

SELECT cast(date_format('2018-06-05 15:25:42.23','yyyy-MM-dd') as date); -- 2018-06-05 DATE_TYPE
SELECT cast(date_format(current_date(),'yyyy-MM-dd') as date); -- 2018-06-05 DATE_TYPE
SELECT cast(date_format(current_timestamp(),'yyyy-MM-dd') as date);  -- 2018-06-05 DATE_TYPE

Convert String/Timestamp/Date to BIGINT_TYPE

SELECT to_unix_timestamp('2018/06/05 15:25:42.23','yyyy/MM/dd HH:mm:ss'); -- 1528205142 BIGINT_TYPE
SELECT to_unix_timestamp(current_date(),'yyyy/MM/dd HH:mm:ss'); -- 1528205000 BIGINT_TYPE
SELECT to_unix_timestamp(current_timestamp(),'yyyy/MM/dd HH:mm:ss'); -- 1528205142 BIGINT_TYPE

Convert String/Timestamp/Date to STRING

SELECT date_format('2018-06-05 15:25:42.23','yyyy-MM-dd'); -- 2018-06-05 STRING_TYPE
SELECT date_format(current_timestamp(),'yyyy-MM-dd'); -- 2018-06-05 STRING_TYPE
SELECT date_format(current_date(),'yyyy-MM-dd'); -- 2018-06-05 STRING_TYPE

Convert BIGINT unixtime to STRING

SELECT to_date(from_unixtime(unixtime,'yyyy/MM/dd HH:mm:ss')); -- 2018-06-05 STRING_TYPE

Convert String to BIGINT unixtime

SELECT unix_timestamp('2018-06-05 15:25:42.23','yyyy-MM-dd') as TIMESTAMP; -- 1528149600 BIGINT_TYPE

Convert String to TIMESTAMP

SELECT cast(unix_timestamp('2018-06-05 15:25:42.23','yyyy-MM-dd') as TIMESTAMP); -- 1528149600 TIMESTAMP_TYPE

Idempotent (String -> String)

SELECT from_unixtime(to_unix_timestamp('2018/06/05 15:25:42.23','yyyy/MM/dd HH:mm:ss')); -- 2018-06-05 15:25:42 STRING_TYPE

Idempotent (Date -> Date)

SELECT cast(current_date() as date); -- 2018-06-26 DATE_TYPE

Current date / timestamp

SELECT current_date(); -- 2018-06-26 DATE_TYPE
SELECT current_timestamp(); -- 2018-06-26 14:03:38.285 TIMESTAMP_TYPE

AFAIK you must reformat your String in ISO format to be able to cast it as a Date:

cast(concat(substr(STR_DMY,7,4), '-',
            substr(STR_DMY,1,2), '-',
            substr(STR_DMY,4,2)
           )
     as date
     ) as DT

To display a Date as a String with specific format, then it's the other way around, unless you have Hive 1.2+ and can use date_format()

=> did you check the documentation by the way?

Let's say you have a column 'birth_day' in your table which is in string format, you should use the following query to filter using birth_day

date_Format(birth_day, 'yyyy-MM-dd')

You can use it in a query in the following way

select * from yourtable
where 
date_Format(birth_day, 'yyyy-MM-dd') = '2019-04-16';

This will convert the whole column:

select from_unixtime(unix_timestamp(transaction_date,'yyyyMMdd')) from table1