I have a date field like this: 2017-03-22 11:09:55 (column name: install_date)
I have another date field with date like this: 2017-04-20 (column name: test_date)
I would like to obtain only the date field from the above (2017-03-22) so that I can perform a DATEDIFF between install_date and test_date.
Assuming you are looking for this in Hive, you can use TO_DATE function.
TO_DATE('2000-01-01 10:20:30') returns '2000-01-01'
NOTE: Input to TO_DATE is a string
Related
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 have a 'timestamp without time zone' value in a column example:
PunchIn = "2019-06-10 09:10:00"
How can I extract "2019-06-10 00:00:00" from this column?
demo:db<>fiddle
SELECT my_date::date::timestamp
This converts the date column my_date to date (cuts the time part) and if you cast it back into timestamp it gets the 0 time.
Alternatively you can use the date_trunc function:
SELECT date_trunc('day', my_date)
SELECT SUBSTRING("PunchIn = '2019-06-10 09:10:00'" FROM 11 FOR 21);
should return the string you're looking for (if I count the index number right:)
You can use the DATE function to extract date:
SELECT DATE(SUBSTRING("PunchIn = '2019-06-10 09:10:00'" FROM 11 FOR 21));
I'm trying to get records inserted after a certain date given to me by the client.
2018-06-06
Here's how I'm writing the query:
{:ok, date} = NaiveDateTime.from_iso8601(date_string)
from(
m in query,
where: m.inserted_at > ^date
)
(MatchError) no match of right hand side value: {:error, :invalid_format}
And when I try to use a simple Date object:
** (Ecto.Query.CastError) lib/messages/search.ex:77: value ~D[2018-06-06] in where cannot be cast to type :naive_datetime in query
How can I find all messages inserted after that dummy string date the client is passing me?
You have an ISO 8601 date there, not a datetime. You can convert it into a NaiveDateTime (with hour, minute, second all set to 0) like this:
iex(1)> date_string = "2018-06-06"
"2018-06-06"
iex(2)> ndt = NaiveDateTime.from_iso8601!(date_string <> " 00:00:00")
~N[2018-06-06 00:00:00]
Now you can use ndt in your query and it will work.
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.
Say I have a column that contains unix timestamps - an int representing the number of seconds since the epoch. They look like this: 1347085827. How do I format this as a human-readable date string in my SELECT query?
Postgresql has a handy built-in function for this: to_timestamp(). Just wrap that function around the column you want:
Select a, b, to_timestamp(date_int) FROM t_tablename