I have a Decimal value in the column. I have Decimal values converted into hours.
example : my column in one value 3600 then my hours' column in 01:00:00 (HH:MM:SS)format.
Second example : value 1800 then output is 00:30:00 (HH:MM:SS) format
Related
I am formatting an input that I get in Timestamp datatype and sending the output as a text in the below format
2020-07-30 10:45:23.638 PM
For this I'm using the below query
select to_char(input_timestamp, 'YYYY-MM-DD HH:MI:SS.MS AM');
But I want the microsecond part truncated to 2 digits instead of 3 digits. The output will be in TEXT datatype. Expecting help on this.
Try this:
select
left(to_char(input_timestamp::timestamp(2), 'YYYY-MM-DD HH:MI:SS.MS'), 22) ||
to_char(input_timestamp, 'FM AM');
First round to two microsecond' digits, convert to string, chop to 22 characters and then concatenate the meridiem indicator.
I have a column called PairDt, a string that contains a date value in the last 5 characters. I want to compare that date value with the date value in the Day column, which contains dates in the YYYY-MM-DD format.
PairDt Day
----------------------------------
DCS-CNY-Yunbi-42606 2016-08-24
DCS-CNY-Yunbi-42607 2016-08-25
DCS-CNY-Yunbi-42608 2016-08-26
DCS-CNY-Yunbi-42609 2016-08-27
DCS-CNY-Yunbi-42610 2016-08-28
How do I convert Day to a value?
I'm trying to isolate Date values in PairDt that does not match the date value in Days
This 5 digit number at the end of PairDt looks like number of days since December 30th 1899. To convert this number to date use DATEADD to add as many days. To convert a date to number, use DATEDIFF to calculate the number of days. Something like this code:
declare #PairDt varchar(50) = 'DCS-CNY-Yunbi-42606', #Day date = '2016-08-24'
select DATEADD(d, cast(right(#PairDt, 5) as int), '1899-12-30'), DATEDIFF(day, '1899-12-30', #Day)
Excel file(104976x10) includes large data.
A column: Time (unit year)
B column: Year
C column: Day of the year
D column: Hour
E column: Minute
and others including values
I would like to convert column which begins with B column until E column to date format like 'dd/mm/yyyy HH:MM'.
Example for the data:
1998,41655251 1998 152 1 0 12,5 12,0 11,8 11,9 12,0
I would like to do date instead of 2-th, 3-th, 4-th and 5-th columns.
1998,41655251 01/06/1998 01:00 12,5 12,0 11,8 11,9 12,0
or
1998,41655251 01/06/1998 01:00 1998 152 1 0 12,5 12,0 11,8 11,9 12,0
Welcome to SO.
Matlab has two types of date-format:
datetime, introduced in 2014b.
datenum, introcuced in long ago (before 2006b), it is basically a double precision value giving the number of days from January 0, 0000.
I think the best way is to use datetime, and give it the year, month, day, hour and minute values like this:
t=datetime(1998,0,152,1,0,0)
t= '01-May-1998 01:00:00'
As you can see the days automatically overflow into the months. But I end up 1st of may, not 1st of june like in your example.
to change the format:
t.Format='dd/MM/yyyy hh:mm'
t= '01/05/1998 01:00'
to convert it to a string, you can simply use string(t)
This is an example that combines the above functions to read an xlsx file and writes a new one with the updated column.
data=xlsread('test.xlsx');
S = size(data);
t = datetime(data(:,2),0,data(:,3),data(:,4),data(:,5),0);
t.Format='dd/MM/yyyy HH:mm';
data2=num2cell(data(:,1));
data2(:,2)=cellstr(string(t));
data2(:,3:S(2)-3)=num2cell(data(:,6:end));
xlswrite('test2.xlsx',data2);
SELECT cu.user_id, cu.last_activity, cu.updated_time,
DATE_PART('day', cu.last_activity - cu.updated_time), to_char(end_date - start_date, 'DD.HH24')
FROM stats.core_users cu
WHERE cu.user_id = '117132014' or cu.user_id = '117132012';
Get the result like:
117132014 2017-12-11 10:34:51.349905 2017-12-09 12:00:38.503518 1 01.22
117132012 2017-12-11 05:18:20.312283 2017-12-08 15:46:51.914085 2 02.13
Is is feasible to get the day difference with fractions like 1.91 days in the first case, instead of 1 days and 22 hours, to be more precise and easier to fit in a machine learning model?
date_part() does what it's name says: it returns one part of several elements from a date, interval or timestamp. In your case it's one part of an interval (because timestamp - timestamp returns an interval).
If you want the result as a fraction, you need to extract the seconds of the interval and then divide that by 86400 (which is the number of seconds in a day)
extract(epoch from cu.last_activity - cu.updated_time) / 86400
Is there a built in function in PostgreSQL 9.5 version to calculate the appropriate century/millenium?
When I use birth_date::TIMESTAMP from a table, sometimes it prefix 19 and sometimes it prefix 20. Below example
Input:
28JUN80
25APR48
Output:
"1980-06-28 00:00:00"
"2048-04-25 00:00:00"
I also have records in the table with birth_date holding values like "07APR1963" which gets computed appropriately as "1963-04-07 00:00:00".
I need use CASE statement when the length is 7 characters, then prefix with 19 millennium and when its 9 characters, just load it as it is.
https://en.wikipedia.org/wiki/Unix_time Unix epoch is
beginning (00:00:00 1 January 1970)
So if you don't specify the century, but just last YY it will be 20th century from 00:00:00 1 January and 21st century before YY equal 70. If you want it to guess the 20th century either append year as you do, or specify CC, eg:
t=> select
to_timestamp('1JAN70', 'ddmonYY')
, to_timestamp('31DEC69', 'ddmonyy')
, to_timestamp('31DEC69 20', 'ddmonyy cc');
to_timestamp | to_timestamp | to_timestamp
------------------------+------------------------+------------------------
1970-01-01 00:00:00+00 | 2069-12-31 00:00:00+00 | 1969-12-31 00:00:00+00
(1 row)
https://www.postgresql.org/docs/current/static/functions-formatting.html
In conversions from string to timestamp or date, the CC (century)
field is ignored if there is a YYY, YYYY or Y,YYY field. If CC is used
with YY or Y then the year is computed as the year in the specified
century. If the century is specified but the year is not, the first
year of the century is assumed.
update
So in your case you should do smth like:
vao=# create table arasu (member_birth_date character(9)); insert into arasu values ('28JUN80'),('25APR48');
CREATE TABLE
INSERT 0 2
vao=# select to_timestamp(member_birth_date||' 20', 'ddmonYY cc') from arasu;
to_timestamp
------------------------
1980-06-28 00:00:00+03
1948-04-25 00:00:00+03
(2 rows)