I have a table in PostgreSQL.
This table has a months column and a mydate column.
The months has a value of Jan the mydate has a value of 2017-01-01
I want to update that value to 2018-01-01 but I don't want to have to do it by hard coding in the 2018 date. I would like to use a date_part function but I am not sure if I am approaching this correctly.
Here is what I have so far it is not complete I am stuck on what I need to finish this query:
UPDATE tblshopstatus
Set mydate = mydate + date_part('year') -----I am stuck on this line----
WHERE months = 'Jan'
More examples:
In the months column I have all 12 months listed.
In the mydate column I have dates listed as 2017-01-01, 2017-02-01 etc... through the 12 months.
Is there a way to just increase the year to 2018 for all months.
set mydate = mydate + interval '1 year'
Related
I am trying to get week numbers in a Year starting from a certain day
I've checked the stack but quite confused.
SELECT EXTRACT(WEEK FROM TIMESTAMP '2021-01-01'),
extract('year' from TIMESTAMP '2021-01-01')
The output is 53|2021
I want it to be 01|2021
I understand the principle of the isoweek but I want the year to start in 01-01-2021
The aim is to use intervals from this day to determine week numbers
Week N0| End Date
1 | 01-01-2021
2 | 01-08-2021
5 | 01-29-2021
...
This is really strange way to determine the week number, but in the end it's a simple math operation: the number of days since January first divided by 7.
You can create a function for this:
create function custom_week(p_input date)
returns int
as
$$
select (p_input - date_trunc('year', p_input)::date) / 7 + 1;
$$
language sql
immutable;
So this:
select date, custom_week(date)
from (
values
(date '2021-01-01'),
(date '2021-01-08'),
(date '2021-01-29')
) as v(date)
yields
date | custom_week
-----------+------------
2021-01-01 | 1
2021-01-08 | 2
2021-01-29 | 5
I have a whole bunch of tariffs, some work on weekends, some work on weekdays some on both. Sometimes I'll be querying on NOW() but sometimes I'll be querying on datetime column.
id | Weekday | Weekend | Price
1 | 1 | 0 | 0.04
2 | 0 | 1 | 0.02
date
2020-04-15 00:00:00
2012-04-16 00:00:00
The date is from another table and is not related to the Price / days of week.
I know I can get the weekend dates by
SELECT * FROM tariff where EXTRACT(ISODOW FROM date) IN (6,7)
however I can't think of how I'd get rows that are either weekend / weekdays or both given a date.
** edit **
Updated the tables to show the dates are seperate. What I'm trying to get is the tariff that corresponds to the date in that table, whether it's on a week day or a weekend (or both but I can extrapolate that).
The weekend 1 is the tariff that is used for weekends, weekdays 1, all days is both.
Cannot give you a query, supply anything to query. Nor can we be sure that the columns Weekday and Weekend mean as you didn't tell us. But if we take them as boolean indicator where 1 means desired may some thing like will work for you.
select ...
from ...
where ...
and ( (weekday = 1 and weekend =1)
or (weekday = 1 and extract(isodow from date) not in (6,7))
or (weekend = 1 and extract(isodow from date) in (6,7))
) ;
I am currently writing a Crystal Report that has a DB2 query as its backend. I have finished the query but am stuck on the date portion of it. I am going to be running it twice a month - once on the 16th, and once on the 1st of the next month. Here's how it should work:
If I run it on the 16th of the month, it will give me results from the 1st of that same month to the 15th of that month.
If I run it on the 1st of the next month, it will give me results from the 16th of the previous month to the last day of the previous month.
This comes down a basic bi-monthly report. I've found plenty of hints to do this in T-SQL, but no efficient ways on how to accomplish this in DB2. I'm having a hard time wrapping my head around the logic to get this to consistently work, taking into account differences in month lengths and such.
There are 2 expressions for start and end date of an interval depending on the report date passed, which you may use in your where clause.
The logic is as follows:
1) If the report date is the 1-st day of a month, then:
DATE_START is 16-th of the previous month
DATE_END is the last day of the previous month
2) Otherwise:
DATE_START is 1-st of the current month
DATE_END is 15-th of the current month
SELECT
REPORT_DATE
, CASE DAY(REPORT_DATE) WHEN 1 THEN REPORT_DATE - 1 MONTH + 15 ELSE REPORT_DATE - DAY(REPORT_DATE) + 1 END AS DATE_START
, CASE DAY(REPORT_DATE) WHEN 1 THEN REPORT_DATE - 1 ELSE REPORT_DATE - DAY(REPORT_DATE) + 15 END AS DATE_END
FROM
(
VALUES
DATE('2020-02-01')
, DATE('2020-02-05')
, DATE('2020-02-16')
) T (REPORT_DATE);
The result is:
|REPORT_DATE|DATE_START|DATE_END |
|-----------|----------|----------|
|2020-02-01 |2020-01-16|2020-01-31|
|2020-02-05 |2020-02-01|2020-02-15|
|2020-02-16 |2020-02-01|2020-02-15|
In Db2 (for Unix, Linux and Windows) it could be a WHERE Condition like
WHERE
(CASE WHEN date_part('days', CURRENT date) > 15 THEN yourdatecolum >= this_month(CURRENT date) AND yourdatecolum < this_month(CURRENT date) + 15 days
ELSE yourdatecolum > this_month(CURRENT date) - 1 month + 15 DAYS AND yourdatecolum < this_month(CURRENT date)
END)
Check out the THIS_MONTH function - there are multiple ways to do it. Also DAYS_TO_END_OF_MONTH might be helpful
I am trying to write a query which gives the number of days in each month between two specified dates.
Example:
date 1: 2018-01-01
date 2: 2018-05-23
Expected Output:
month days
2018-01-01, 31
2018-02-01, 28
2018-03-01, 31
2018-04-01, 30
2018-05-01, 23
Use generate_series and group by date_trunc
SELECT date_trunc('month',dt) AS month,
COUNT(*) as days
FROM generate_series( DATE '2018-01-01',DATE '2018-05-23',interval '1 DAY' )
as dt group by date_trunc('month',dt)
order by month;
Demo
I have this field named late_in that contains data like this 2017-05-29 08:36:44 where the limit for entry time is 08:30:00 every day.
What I want to do is to get the year, month and how many times he late in that month even if it zero late in the month.
I want the result look something like this:
year month late
-------------------
2017 1 6
2017 2 0
2017 3 14
and continue until the end of year.
You are looking for conditional aggregation:
select extract(year from late_in) as year,
extract(month from late_in ) as month,
count(*) filter (where late_in::time > time '08:30:00') as late
from the_table
group by extract(year from late_in),
extract(month from late_in );
This assumes that late_in is defined as timestamp.
The expression late_in::time returns only the time part of the value and the filter() clause for the aggregation will result in only those rows being counted where the condition is true, i.e. where the time part is after 08:30