Here is my 1 line of data (for brevity):
73831 12/26/2014 1:00:00 AM 0.3220
The 2nd column is the time column which is in string format. I'm using this hive query:
select col2, UNIX_TIMESTAMP(col2,'MM/DD/YYYY hh:mm:ss aaa') from Table
Here is what I get: 1388296800
However, when I check with, http://www.epochconverter.com/ and also from_unixtime(1388296800), I get a different date.
Is there something wrong with my format / pattern string I enter into UNIX_TIMESTAMP in Hive?
Your date format symbols need to conform to those in the Java SimpleDateFormat documentation.
For your date it looks like you want MM/dd/yyyy HH:mm:ss aa.
Related
I'm trying to cast the results of an entire column in BigQuery in MM/DD/YYYY format
Time_Column
2022-05-26T17:32:41.000Z
2022-05-28T06:34:23.000Z
Results:
We can try to use FORMAT_DATE function to make it, we can refer to this link Supported Format Elements For DATE to use your expected date format from the datetime type.
SELECT FORMAT_DATE("%m/%d/%Y", Time_Column)
FROM T
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')
I have a table with the following data:
logs.ip logs.fecha logs.metodo
66.249.93.79 19/Nov/2018:03:46:33 GET
All data columns are string and I want to convert logs.fecha into date with the following format: YYYY-MM-dd HH:mm:ss
I try the following query:
SELECT TO_DATE(from_unixtime(UNIX_TIMESTAMP(fecha, 'yyyy-MM-dd'))) FROM logs
Results of the query are NULL in all rows.
How can I make the conversion string to date for all rows? I know I must use ALTER TABLE but I don't know how to do it.
Thanks
The reason you get null is because the format of the input string is different from the input passed to unix_timestamp. The second argument to unix_timestamp should specify the string format of the first argument. In from_unixtime you can specify the output format desired. If nothing is specified, a valid input to from_unixtime returns an output in yyyy-MM-dd format.
The error can be fixed as below.
from_unixtime(unix_timestamp(fecha,'dd/MMM/yyyy:HH:mm:ss'),'yyyy-MM-dd HH:mm:ss')
You just have to tell Oracle the date format you are reading with TO_DATE.
Try:
SELECT TO_DATE(fecha,'DD/MON/YYYY:HH:MI:SS') FROM logs
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'm new to sql loader, and i'm having a problem with the date format.
Here is the input file record sample:
Y,1525039510,http-192.168.2.2,15-01-2011 00:00:032:728,64
Y,1525131958,http-192.168.2.2,15-01-2011 00:00:033:613,75
I'm having a problem with the fourth column, the date.
My current field entry is this:
start_time DATE "DD-MM-YYYY HH:MI
How do i parse the last part of the input 032:728 (seconds and milliseconds)
I tried SSS:FF3 and SS.FF3, no luck.
You could treat the input as a string and transform it with a function. Something like this should work:
start_time CHAR to_date(substr(:start_time, 1, 20), 'dd-mm-yyyy hh24:mi":0"ss')
I had to drop the millisecond part since they are not part of the DATE datatype (to_date doesn't recognize the FF format).
You can read this into a timestamp field. The data looks a bit weird: 15-01-2011 00:00:033:613,75 but the conversion format should be dd-mm-yyyy hh[24]:mi:0ss:fff,ff
Make sure that your nls settings are as expected.
Ronald.