I have a date field in a postgresql database (field name is input) how can I extract the month only from the date field? I used the syntax below, but I want it to show the actual month name, not a numeric value for the month
EXTRACT(MONTH FROM input) AS "Month"
So if the date in the field input was 04/26/2016 this syntax returns 4, how could I alter this to return April
A simpler version.
to_char(input, 'Month') AS Month
Source : Data Type Formatting Functions
You may find this useful
SELECT to_char(to_timestamp (date_part('month', timestamp '2016-04-26 20:38:40')::text, 'MM'), 'Month')
All the best
Ref Postgresql (Current) Section 9.9. Date/Time Functions and Operators
Related
In a PostgreSQL table I have two columns (int8) containing a Unix timestamp. [Note: this table is not mine, so the columns can be changed to an actual date, hence the conversion in the query.]
below is a stripped down version of my query. What I need to do is to group the entries by month. But in my query below, the entries november 2020 get grouped with the entries of november 2021.
select
DATE_PART('month',to_timestamp(e.startts)) as "Date", sum(e.checked)
from entries e
where
e.startts >= date_part('epoch', '2020-10-01T15:01:50.859Z'::timestamp)::int8
and e.stopts < date_part('epoch', '2021-11-08T15:01:50.859Z'::timestamp)::int8
group by "Date"
How can I have "Date" as month/year instead of month? so for example 10/20 instead of 10.
Use to_char() for format the timestamp value:
to_char(to_timestamp(e.startts), 'mm/yy') as "Date"
I am trying to extract a month from a date in SAS, but so far all my new month variables are coming up as missing.
I have attempted to use some combinations of the month() function in SAS, but so far it just comes up as missing. The dates are formatted as follows: 01/31/2017 (MMDDYY10.)
I have tried
month = month(end_date)
Month =catx('/',put(month(end_date),z2
I would like the Month to show up as a number (01) or a 3 letter code (JAN), currently it is just missing (.)
Thanks in advance!
For month() to return a missing value the end_date variable must be numeric and missing. If end_date were a character variable the log would show invalid numeric data.
Use the monname3. format to convert a date value to a $3. character value mon
monthname = put (end_date, monname3.);
Other alternatives are:
keep the date value unchanged and change the format, or
map the date value to the first of the month value and also format that
For example:
end_date_copy = end_date;
format end_date_copy monname3.;
end_date_month = intnx('month', end_date, 0);
format end_date_month monname3.;
What you ultimately do depends on how the mon is to be used downstream in reporting or aggregating.
I'm trying to add only the current date in "DD/MM/YYYY" format in a field of type ' timestamp in PostgreSQL' .
I try:
select to_char(now(),'DD/MM/YYYY') as date;
But PostgreSQL return me:
TIP : You will need to rewrite the expression or apply a type conversion.
There is no such thing as "only the date" in a timestamp field. A timestamp field will store timestamps.
Try using the date type instead. Please read about this here.
Also, please consider using the ISO 8601 format instead. Getting used to it helps in a lot of cases.
Human-readable formats like "DD/MM/YYYY" should only be used for presentation.
If you want to use timestamp fields and insert a human-readable formatted dates, then you are looking for:
to_timestamp('05/04/2016', 'DD/MM/YYYY')
If it is about the current date, then Postgres provides the CURRENT_DATE function, which you may use:
SELECT CURRENT_DATE;
INSERT INTO t (timestamp_field) VALUES (CURRENT_DATE);
According to the PostgreSQL version 8.1 date-time function docs:
dow
The day of the week (0 - 6; Sunday is 0) (for timestamp values only)
SELECT EXTRACT(DOW FROM TIMESTAMP '2001-02-16 20:38:40');
Result: 5
Why is not a Date datatype also a valid argument for this function? If the sequence of the days of the week does not change by locale:
0 - 6; Sunday is 0
why would the time-component of a combined date-type value be needed to determine the ordinal of the day in the week? Wouldn't the date-chunk alone be sufficient?
The original question referenced version 8.1, the 9.5 documentation states:
EXTRACT(field FROM source)
The extract function retrieves subfields such as year or hour from date/time values. source must be a value expression of type timestamp, time, or interval. (Expressions of type date are cast to timestamp and can therefore be used as well.) field is an identifier or string that selects what field to extract from the source value. The extract function returns values of type double precision. The following are valid field names:
And then later, specifically under dow:
dow
The day of the week as Sunday (0) to Saturday (6)
SELECT EXTRACT(DOW FROM TIMESTAMP '2001-02-16 20:38:40');
Result: 5
Note that extract's day of the week numbering differs from that of the to_char(..., 'D') function.
This is just a documentation bug.
I have create one field in sql server database as nvarchar datatype and store some date like 'd/MM/yyyy' and 'dd/MM/yyyy' format previously. Now i want to get all data in 'dd/MM/yyyy' format using query it is possible?
You can cast the field to datetime in the query:
select cast(YourField as datetime)
from YourTable
where isdate(YourField) = 1
The where isdate(YourField) = 1 part is necessary to filter out rows where the value is no valid date (it's a nvarchar field, so there could be things like abc in some rows!)
But you should really change the field to datetime in the long term, as already suggested by Christopher in his comment.
Casting like described above is always error-prone because of the many different data formats in different countries.
For example, I live in Germany where the official date format is dd.mm.yyyy.
So today (December 9th) is 9.12.2011, and running select cast('9.12.2011' as datetime) on my machine returns the correct datetime value.
Another common format is mm/dd/yyyy, so December 9th would be 12/9/2011.
Now imagine I have a nvarchar field with a date in this format on my German machine:
select cast('12/9/2011' as datetime) will return September 12th (instead of December 9th)!
Issues like this can easily be avoided by using the proper type for the column, in this case datetime.