I have timestamp column but the column is in string datatype but it contains actual date value in hexadecimal format as below. How to convert this hexadecimal date value into date value using hive
0x0000003D02E2CE75
0x0000000945EB44AE
0x0000000945EB44B0
0x0000000945EB44D3
0x00000008A11F0E9A
0x00000008A11F0ECF
you can use this (use regexp_replace instead of replace for hive 1.1):
select
from_utc_timestamp(cast(conv(replace('0x00000008A11F0ECF', 'x', '0'), 16, 10) as bigint), 'PST')
Related
I have a date time field which is coming from an external system in UTC format 2022-01-02T08:00:00.000+00:00. This value should be queried in DB2 to determine whether the record exists or not. The date stored in DB2 is in the format 2022-01-01 08:00:00.000 Is there any way to convert the incoming date in the format 2022-01-01 08:00:00.000 ?
The final query should be something like
select * from table where changedate = '2022-01-02T08:00:00.000+00:00'
Db2 doesn't store timestamps in a string format. Some binary format is used for that.
So, if I got you right, your question should be changed to "how to convert YYYY-MM-DDTHH24.MI.SS.FF3XXXXXX string representation of timestamp to the timestamp data type".
Unfortunately, there is no such a built-in pattern in the TO_DATE / TIMESTAMP_FORMAT function, but you can use the following expression. T column has the timestamp data type, and you may use this expression in the select * from table where changedate = ... statement.
SELECT
S
, TO_DATE (TRANSLATE (SUBSTR (S, 1, 23), ' ', 'T'), 'YYYY-MM-DD HH24:MI:SS.FF3')
+ CAST (TRANSLATE (SUBSTR (S, 24), '', ':', '') || '00' AS DEC (6)) AS T
FROM
(
VALUES
'2022-01-02T08:00:00.000+00:00'
, '2022-01-02T08:00:00.000+03:30'
, '2022-01-02T08:00:00.000-03:30'
) T (S)
S
T
2022-01-02T08:00:00.000+00:00
2022-01-02-08.00.00.000000
2022-01-02T08:00:00.000+03:30
2022-01-02-11.30.00.000000
2022-01-02T08:00:00.000-03:30
2022-01-02-04.30.00.000000
Below is a script i am trying to run in Presto; Subtracting today's date from an integer field I am attempting to convert to date. To get the exacts days between. Unfortunately, it seems the highlighted block does not always convert the date correctly and my final answer is not correct. Please does anyone know another way around this or a standard method on presto of converting integer values to date.
Interger value in the column is in the format '20191123' for year-month-date
select ms, activ_dt, current_date, date_diff('day',act_dt,current_date) from
(
select ms,activ_dt, **CAST(parse_datetime(CAST(activ_dt AS varchar), 'YYYYMMDD') AS date) as act_dt**, nov19
from h.A_Subs_1 where msisdn_key=23480320012
) limit 19
You can convert "date as a number" (eg. 20180527 for May 27, 2018) using the following:
cast to varchar
parse_datetime with appropriate format
cast to date (since parse_datetime returns a timestamp)
Example:
presto> SELECT CAST(parse_datetime(CAST(20180527 AS varchar), 'yyyyMMdd') AS date);
_col0
------------
2018-05-27
You can use below sample query for your requirement:
select date_diff('day', date_parse('20191209', '%Y%m%d'), current_timestamp);
I'm trying to convert value for DIM_DT_ID to MMddYY. I'm successful in doinf that. However, query fails because ultimately I'm comparing a character value to date here. Is there a way by which I can get value for DIM_DT_ID in MMddyy format and its data type still remains DATE ?
Here DIM_DT_ID
SELECT DIM_DT_ID
DIM_DT_ID >= FORMATDATE('MMddyy',ADDDAY(TO_date('yyyy-MM-dd','2016-12-21'), -25)); from abc;
Regards,
Ajay
In Denodo, to convert a string to a date field, use "to_date()" (which returns a date).
Then, don't convert back to a string, leave that field as a date (so don't use "Formatdate()", which returns a string).
So:
SELECT *
FROM MyTable
WHERE now() >= to_date('yyyy-MM-dd',myStringFieldThatLooksLikeADate)
In my example, "now()" is a date, and so is the output of the to_date function... so you can do a comparison.
If you try to convert the date back to a string using formatdate, it won't work:
#This doesn't work:
SELECT *
FROM MyTable
WHERE now() >= formatdate('MMddyy',to_date('yyyy-MM-dd',myStringFieldThatLooksLikeADate))
It doesn't work because we are comparing a date ("now()") to a string.
I have a field that should be 6 digit character but it is numeric. I am using the following code to add the leading zero:
select CAST(CAST(CHD_OPEN_DATE AS FORMAT '9(6)') AS CHAR(9))
I'm using the following code to format this as a date:
cast(cast(lpad(to_char(CHD_OPEN_DATE),6,'0') as date format 'YYMMDD') as date format 'YYYY-MM-DD')
When using this date format 1990 comes up as 2090. Is there a work-around for this?
If your number has a YYMMDD format you can use the following to cast to a date without the need to cast to an intermediate string. Assuming a date range between 1930 and 2029:
SELECT 900331 AS CHD_OPEN_DATE,
Cast(CASE WHEN CHD_OPEN_DATE < 300000
THEN CHD_OPEN_DATE + 1000000
ELSE CHD_OPEN_DATE
END AS DATE)
I have a CHAR_Date column containing date values in the format 'YYMMDD'.
I would like to do date arithmetic so I need to convert it into a Date data type. The problem is that the Char_Date also contains Blanks.
How do I cast the CHAR_Date to a DATE_Date column, with valid values?
SELECT
case when CHAR_Date = '' then TIMESTAMP('0001-01-01')
else TIMESTAMP_FORMAT(CHAR_Date, 'YYMMDD')
end
as DATE_Date
FROM TABLE_Data
You can use the function TIMESTAMP_FORMAT
TIMESTAMP_FORMAT('990205' , 'YYMMDD')
And if you want a date:
DATE(TIMESTAMP_FORMAT('990205' , 'YYMMDD'))