I want to show date in string format as current month name(String) and year(e.g -> SEPTEMBER, 2019). How to write query for this in PostgreSQL?
I tried this query:
select
date(date_trunc('month', current_date));
but it gives me only current months starting date.
Try to_char() and add the year formatter to the string like this:
SELECT to_char(current_date, 'MONTH YYYY')
This will return:
SEPTEMBER 2019
Here's a sqlfiddle
If you want to format the output of a DATE value, use to_char()
select to_char(current_date, 'Month, yyyy');
we can do it like this also in Postgresql -
SELECT TO_CHAR(NOW() :: DATE, 'Month , yyyy');
output - September, 2019
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 SELECT statement that I am trying to convert from Netezza SQL to Impala SQL. The output looks something like 140612, which is a date that is obtained by subtracting 7 from the current date and then pulling out the monday of that week.
I need to have this readable for Impala, then format it, then turn it into a string.
The query is :
TO_CHAR(next_day(DATE(a.date)-7, 'Monday'), 'YYMMDD') AS START_DATE
Assuming a.date is a timestamp, and T is the day of the week (1 = Sunday, 7 = Saturday; for your example above, Monday = 2, so T = 2) you should be able to use use
date_add(a.date, 7 - pmod(dayofweek(a.date) - T, 7));
in place of next_day in the above query. Check out the documentation on Impala's built-in date and time functions for more detail.
How do I return a character string that represents the specified datepart of the specified date using PostgreSQL?
Note: In SQL Server we use the following syntax.
Example: (Using SQL Server)
SELECT DATENAME(DW,'2014-05-22'); /*For Weekday */
As documented in the manual the to_char() function can be used for this:
select to_char(DATE '2014-05-22', 'Day');
returns Thursday
date_part(text, timestamp)
Example for day of the week:
select date_part('dow', timestamp '2014-05-22 00:00:00')
I have a query that produces date in the datetime format... see below.
UPDATE WCV_3
SET VISIT_1= #TEMP2.[FIRST SERVICE DATE]
FROM WCV_3
JOIN #TEMP2 ON #TEMP2.UNIVERSALMEMBERID= WCV_3.UNIVERSALMEMBERID
WHERE #TEMP2.VISIT_CNT= 1
returns this: 2013-01-03 00:00:00.000
What is the correct CONVERT statement to use and where do I place it in this query to return the DATE only and not the Time part?
I want just 2013/01/03
UPDATE WCV_3
SET VISIT_1= cast(#TEMP2.[FIRST SERVICE DATE] as date)
FROM WCV_3
JOIN #TEMP2 ON #TEMP2.UNIVERSALMEMBERID= WCV_3.UNIVERSALMEMBERID
WHERE #TEMP2.VISIT_CNT= 1
You can use a cast to change the datetime to just a date if your using sql server 2008 +
Select Cast(getdate() as date)
I have a column in a table in which we are storing date in DATETIME format. (DD-MON-RRRR HH24:MI:SS) - Database Oracle 11g.
Data Type of a column is DATE, and storing date in 01-01-2012 01:00 PM (i.e. jan 1, 2012) format.
entity
#NotNull
#Column(name = "dateColumnName")
#Temporal(TemporalType.TIMESTAMP)
private Date sampleDate;
I am fetching all data by passing date
SAMPLE_QUERY = "select * from TableA tab where tab.dateWithTime = :sampleDate order by tab.dateWithTime ASC "
singleDate is "Tue Jan 24 00:00:00 IST 2012" , fasttime :
1327343400000
The problem is I am passing only date in the query, though Date through which records are being fetched is in DATE TIME format i.e 01-01-2012 01:00 PM.
How can i change my query so that it fetches all the records in ascending order of DateTime.
If you want to fetch all times for that day, then change your query to be more like
SELECT ... WHERE dateField >= :lowerParam AND dateField < :upperParam
Oracle has no DATE TIME datatype. The DATE datatype contains both a date and a time component, down to the second. TIMESTAMPS get a bit more complicated.
If your dateWithTime column is indeed a DATE datatype, the ORDER BY dateWithTime ASC clause should order your results in ascending order.
You may not be displaying the time component of your date. You can convert a date to a string in that format with TO_CHAR( dateWithTime, 'dd/mm/yyyy hh24/mm/ss' ) or whatever format you want.
Edit:
Oh, do you mean you want to find the cases where the date component of the DATE matches, but you don't care about the time component? That can be handled in the where clause with something like:
WHERE TRUNC( tab.dateWithTime ) = TRUNC( :sampledate )
TRUNC by default truncates a date to the beginning of the day.