I have this date format in hive: 20180618151752
Make the following query to leave it in the following way:
select
concat(substr(xdr.time_stop,1,8),' ',
substr(xdr.time_stop,9,2),':',
substr(xdr.time_stop,10,2),':',
substr(xdr.time_stop,11,2)) as date
from padl.fraude_vozm_xdr;
20180618 18:17:52
Now, I need to convert that string field to date, how could I do it?.
Thank you
Use from_unixtime and unix_timestamp.
from_unixtime(unix_timestamp('20180618151752','yyyyMMddHHmmss'),'yyyyMMdd HH:mm:ss')
Related
Trying CONVERT()
Trying STR_TO_DATE()
Trying CAST() - neither DATE or DATETIME work
Trying CAST(STR_TO_DATE())
Hi, I cannot figure out what I'm doing wrong with trying to convert my varchar column Procedure_Date containing a string in the dd/mm/YYYY hh:mm:ss format. I am trying to extract just the date from this string but I seem to be cursed. Using MariaDB and have gone through all the docs there. Greatly appreciate any advice!
You can use cast
SELECT Procedure_Date, CAST(STR_TO_DATE(Procedure_Date, '%d-%m-%Y') AS DATE) FROM uniTable5;
Use: DATE('YOUR DATETIME')
EXAMPLE: DATE('2022-04-25 14:50:01')
OUTPUT: 2022-04-25. Result in DATE.
For example column name Start_date is having value like below in hive which datatype is string
02-JUN-22 11.13.22 AM CST
I want to convert the value as below
2022-06-02
tried to_Date function but getting null values
You can try below one,
And also if you have 'AM CST' values included as part of your input string, then you can trim or take substring value to match the below query.
select from_unixtime(unix_timestamp('02-JUN-22 11.13.22' ,'dd-MMM-yy'), 'yyyy-MM-dd');
Use the built-in DateTime/TimeZone functionality
<?php
$mysqlDate = '2009-04-01 15:36:13';
$dateTime = new DateTime ($mysqlDate);
$dateTime->setTimezone(new DateTimeZone('America/Los_Angeles'));
?>
have you tried to_char on your query?
for example:
select to_char(start_date,'yyyy-mm-dd') from your table;
Convert to timestamp in Hive format taking into account the timezone (use from_unixtime(unix_timestamp(col, pattern))), see patterns, then use to_date.
Demo:
select to_date(from_unixtime(unix_timestamp('02-JUN-22 11.13.22 AM CST','dd-MMM-yy hh.mm.ss a z')))
Result:
2022-06-02
Based on the example mentioned here in the Snowflake documentation, why are the date and timestamp values returning different values just by changing the ORDER BY clause? Also, I am trying to convert a string to a date format which is not returning correct results in Snowflake while this works fine in other SQL based Engines. Need help from experts on this.
This query
SELECT '20200710', TO_DATE('20200710');
is returning the following output
20200710 | 1970-08-22
Also tried:
SELECT TO_DATE('20200710', 'YYYY-MM-DD');
and got the error:
Can't parse '20200710' as date with format 'YYYY-MM-DD'
To convert to a date data type, you would use:
SELECT TO_DATE('20200710', 'YYYYMMDD')
I would recommend just keeping the date data type. But if you want a string in the format YYYY-MM-DD:
SELECT TO_CHAR(TO_DATE('20200710', 'YYYYMMDD'), 'YYYY-MM-DD')
My table in hive has a filed of date in the format of '2016/06/01'. but i find that it is not in harmory with the format of '2016-06-01'.
They can not compare for instance.
Both of them are string .
So I want to know how to make them in harmory and can compare them. Or on the other hand, how to change the '2016/06/01' to '2016-06-01' so that them can compare.
Many thanks.
To convert date string from one format to another you have to use two date function of hive
unix_timestamp(string date, string pattern) convert time string
with given pattern to unix time stamp (in seconds), return 0 if
fail.
from_unixtime(bigint unixtime[, string format]) converts the
number of seconds from unix epoch (1970-01-01 00:00:00 UTC) to a
string representing the timestamp of that moment in the current
system time zone.
Using above two function you can achieve your desired result.
The sample input and output can be seen from below image:
The final query is
select from_unixtime(unix_timestamp('2016/06/01','yyyy/MM/dd'),'yyyy-MM-dd') from table1;
where table1 is the table name present in my hive database.
I hope this help you!!!
Let's say you have a column 'birth_day' in your table which is in your format,
you should use the following query to convert birth_day into the required format.
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';
Use :
unix_timestamp(DATE_COLUMN, string pattern)
The above command would help convert the date to unix timestamp format which you may format as you want using the Simple Date Function.
Date Function
cast(to_date(from_unixtime(unix_timestamp(yourdate , 'MM-dd-yyyy'))) as date)
here is my solution (for string to real Date type):
select to_date(replace('2000/01/01', '/', '-')) as dt ;
ps:to_date() returns Date type, this feature needs Hive 2.1+; before 2.1, it returns String.
ps2: hive to_date() function or date_format() function , or even cast() function, cannot regonise the 'yyyy/MM/dd' or 'yyyymmdd' format, which I think is so sad, and make me a little crazy.
I am trying to convert from one date format to another. I am not sure how to write the functions.
My source date looks like 01/15/2009 01:23:15
The format I need is 01152009.
Thanks
Try this.
TO_CHAR(TO_DATE('01/15/2009 01:23:15','MM/DD/YYYY HH:MI:SS'),'MMDDYYY')
More info here,
http://psoug.org/reference/date_func.html
Does this work for you? It assumes the date is in date format but will work with timestamp
select to_char(YourDateField,'DDMMYYYY') from dual;
You can always convert it back to a date using the TO_DATE function if you need that format.
select TO_CHAR(TO_DATE('01/15/2009 01:23:15','MM/DD/YYYY MI:HH:SS'),'MMDDYYYY') from dual
if your field is already of data type date then you should only do:
select TO_CHAR(<fieldname>,'ddmmyyyy') ...