Related
These are the summary of issue:
There is an issue of not going inside the if clause and data not inserted.
Need to compute the target_progress_date as the difference between the end_date and start_date and all the interval months to be computed and stored in the variable so as to insert into the project_target_progress table.
Viz: Start date :1-1-2021, End date : 1-12-2021 Months: 1-1-2021
1-2-2021 1-3-2021 .... 12-12-2021
// To check if any values are there
$query_check = pg_query(DBCON,"select id , (COALESCE(revised_cost,administrative_sanction_project_cost,award_tender_project_cost) as cost, COALESCE(award_tender_contract_date,administrative_sanction_date,actual_start_date) as start_date, COALESCE(revised_completion_date,scheduled_completion_date)::date) as end_date from sp_index_v4 where id = $proj_id ");
// If it has no null values then execute the logic - Autoset of target progress before editing it.
if ($query_check != null) {
console.log ('Hi'+$query_check);
$query_cost = pg_query(DBCON,"select id ,(COALESCE(revised_cost,administrative_sanction_project_cost,award_tender_project_cost)/ datediff('month',COALESCE(award_tender_contract_date,administrative_sanction_date,actual_start_date)::date, COALESCE(revised_completion_date,scheduled_completion_date)::date)) as cost_of_one_month, TRUNC(((COALESCE(revised_cost,administrative_sanction_project_cost,award_tender_project_cost)/ datediff('month',COALESCE(award_tender_contract_date,administrative_sanction_date,actual_start_date)::date, COALESCE(revised_completion_date,scheduled_completion_date)::date))/ (COALESCE(revised_cost,administrative_sanction_project_cost,award_tender_project_cost)))*100,2) as cost_percent from sp_index_v4 id = $proj_id");
// Insert for new project entry update for already having project
$ins_query = "insert into project_target_progress(project_id,cum_financial_progress_in_cr,cum_financial_progress_in_percent,cum_physical_progress_in_percent,target_progress_date)values('$id',cost_of_one_month,cost_percent,cost_percent,start_date)";
$insert_data = pg_query(DBCON, $ins_query);
}
else {
echo "Value cannot be inserted";
console.log("inside the else block");
}
I wrote my own datediff function with different opportunities:
create or replace function datediff
(
units varchar(30),
start_t timestamp,
end_t timestamp
)
returns int as $$
declare
diff_interval interval;
diff int = 0;
years_diff int = 0;
begin
if units in ('yy', 'yyyy', 'year', 'mm', 'm', 'month') then
years_diff = date_part('year', end_t) - date_part('year', start_t);
if units in ('yy', 'yyyy', 'year') then
return years_diff;
else
return years_diff * 12 + (date_part('month', end_t) - date_part('month', start_t));
end if;
end if;
diff_interval = end_t - start_t;
diff = diff + date_part('day', diff_interval);
if units in ('wk', 'ww', 'week') then
diff = diff/7;
return diff;
end if;
if units in ('dd', 'd', 'day') then
return diff;
end if;
diff = diff * 24 + date_part('hour', diff_interval);
if units in ('hh', 'hour') then
return diff;
end if;
diff = diff * 60 + date_part('minute', diff_interval);
if units in ('mi', 'n', 'minute') then
return diff;
end if;
diff = diff * 60 + date_part('second', diff_interval);
return diff;
end;
$$ language plpgsql;
Now, we can write a sample query using our function:
select
mm.start_date,
mm.end_date,
test.datediff('year', mm.start_date, mm.end_date) as diff_year,
test.datediff('month', mm.start_date, mm.end_date) as diff_month,
test.datediff('day', mm.start_date, mm.end_date) as diff_day
from (
select '2021-11-25 02:20:54.200'::timestamp as start_date, now()::timestamp as end_date
) mm
Result:
start_date
end_date
diff_year
diff_month
diff_day
2021-11-25 02:20:54.200
2022-02-24 06:10:52.258
1
3
91
I'd like to get reference date as previous saturday when the query is executed before wednesday,after that reference date is executed day.and I want to reuse this reference date as reference_date . to achieve this, I write following queries.But when I try to assign date to reference_date it returned errors.
Are there any way to achieve this ?
Thanks
select into reference_date
case
when date_part(dw, current_date) <= 3 then (
current_date - (
date_part(dw, current_date) :: integer + 1
)
)
else current_date
end;
select reference_date;
Try this :
select
case
when date_part('dow', current_date) <= 3
then current_date - (date_part('dow', current_date) :: integer + 1)
else current_date
end
into reference_date ;
This is my current implementation
SELECT
date_trunc('month', do_date::date)::date as starting_of_the_month,
(date_trunc('month', do_date::date) + interval '1 month' - interval '1 day')::date as ending_of_the_month,
case when 1 + FLOOR((EXTRACT(DAY FROM do_date) - 1) / 7) = 1
THEN date_trunc('week', do_date)::date || ' - ' ||
(date_trunc('week', do_date) + '6 days') ::date end as week1,
case when 1 + FLOOR((EXTRACT(DAY FROM do_date) - 1) / 7) = 2
THEN date_trunc('week', do_date)::date || ' - ' ||
(date_trunc('week', do_date) + '6 days') ::date end as week2,
case when 1 + FLOOR((EXTRACT(DAY FROM do_date) - 1) / 7) = 3
THEN date_trunc('week', do_date)::date || ' - ' ||
(date_trunc('week', do_date) + '6 days') ::date end as week3,
case when 1 + FLOOR((EXTRACT(DAY FROM do_date) - 1) / 7) = 4
THEN date_trunc('week', do_date)::date || ' - ' ||
(date_trunc('week', do_date) + '6 days') ::date end as week4,
case when 1 + FLOOR((EXTRACT(DAY FROM do_date) - 1) / 7) = 5
THEN date_trunc('week', do_date)::date || ' - ' ||
(date_trunc('week', do_date) + '6 days') ::date end as week5
FROM sales_dos
WHERE date_trunc('month', do_date::date)::date >= '2021-02-01' AND date_trunc('month', do_date::date)::date < '2021-02-28'
This is my output for now :
I want the output to display as below :
Week 1 : 2021-02-01 - 2021-02-07
Week 2 : 2021-02-08 - 2021-02-14
Week 3 : 2021-02-15 - 2021-02-21
Week 4 : 2021-02-22 - 2021-02-28
Week 5 : -
Here is another way to do it (example for January 2021).
with
t as (select date_trunc('month', '2021-03-11'::date) as aday), -- any date in Jan-2021
s as
(
select d::date, d::date + 6 ed, extract('isodow' from d) wd
from t, generate_series (aday, aday + interval '1 month - 1 day', interval '1 day') d
)
select format ('Week %s', extract(day from d)::integer / 7 + 1) as weekname, d, ed
from s
where wd = 1;
So what you are looking for is a hybrid ISO with standard Calendar. You are taking the ISO week starting and ending period, but instead of all weeks being exactly 7 days you potentially truncate the 1st and/or last weeks.
The change to need for this is not actually extensive. For initial query returns the in the ISO week begin date instead of the 1st of the month. Then the main query then checks for week 1 and if so produces the 1st of the month. The only twist is determining the ISO week begin date. For this I've just included a function I have had for some time specifically for that. The change to the week_days function are marked --<<<.
create or replace function iso_first_of_week(date_in date)
returns date
language sql
immutable strict
/*
Given a date return the 1st day of the week according to ISO-8601.
I.e. Return the Date if it is Monday otherwise return the preceding Monday
*/
AS $$
with wk_adj(l_days) as (values (array[0,1,2,3,4,5,6]))
select date_in - l_days[ extract (isodow from date_in)::integer ]
from wk_adj;
$$;
create or replace
function week_dates( do_date_in date)
returns table (week_num integer, first_date date, last_date date)
language sql
immutable strict
as $$
with recursive date_list(week_num,first_date,terminate_date) as
( select 1
, iso_first_of_week(do_date_in)::timestamp --<<<
, (date_trunc('month', do_date_in) + interval '1 month' - interval '1 day')::timestamp
union all
select week_num+1, (first_date+interval '7 day'), terminate_date
from date_list
where first_date+interval '6 day' < terminate_date::timestamp
)
select week_num
, case when week_num = 1 --<<<
then date_trunc('month', do_date_in)::date --<<<
else first_date::date --<<<
end --<<<
, case when (first_date+interval '6 day')::date > terminate_date
then terminate_date::date
else (first_date+interval '6 day')::date
end last_date
from date_list;
$$;
---------- Original Reply
You can use a recursive query CTE to get the week number and first date for each week of the month specified. The main query calculates the ending date, shorting the last if necessary. Then wrap that into a SQL function to return the week number and date range for each week. See example.
create or replace
function week_dates( do_date_in date)
returns table (ween_num integer, first_date date, last_date date)
language sql
immutable strict
as $$
with recursive date_list(week_num,first_date,terminate_date) as
( select 1
, date_trunc('month', do_date_in)::timestamp
, (date_trunc('month', do_date_in) + interval '1 month' - interval '1 day')::timestamp
union all
select week_num+1, (first_date+interval '7 day'), terminate_date
from date_list
where first_date+interval '6 day' < terminate_date::timestamp
)
select week_num
, first_date::date
, case when (first_date+interval '6 day')::date > terminate_date
then terminate_date::date
else (first_date+interval '6 day')::date
end last_date
from date_list;
$$;
Response to: "How can i put the output in a single row with week1, week2, week3, week4 and week5". This is essentially the initial output that did not satisfy what you wanted. The term for this type action is PIVOT and is generally understood. It stems from transforming row orientation to column orientation. It is not overly difficult but it is messy.
IMHO this is something that belongs in the presentation layer and is not suitable for SQL. After all you are rearranging the data structure for presentation purposes. Let the database server use its natural format, use the presentation layer to reformat. This allows reuse of the queries instead of rewriting when the presentation is changed or another view of the same data is required.
If you actually want this then just use your initial query, or see the answer from
#Bohemian. However the below shows how this issue can be handled with just SQL (assuming the function week_dates was created).
select week1s
, case when week5e is null
then week4e
else week5e
end "end of month"
, week1s || ' - ' || week1e
, week2s || ' - ' || week2e
, week3s || ' - ' || week3e
, week4s || ' - ' || week4e
, week5s || ' - ' || week5e
from ( select max(case when (week_num=1) then first_date else NULL end) as week1s
, max(case when (week_num=1) then last_date else NULL end) as week1e
, max(case when (week_num=2) then first_date else NULL end) as week2s
, max(case when (week_num=2) then last_date else NULL end) as week2e
, max(case when (week_num=3) then first_date else NULL end) as week3s
, max(case when (week_num=3) then last_date else NULL end) as week3e
, max(case when (week_num=4) then first_date else NULL end) as week4s
, max(case when (week_num=4) then last_date else NULL end) as week4e
, max(case when (week_num=5) then first_date else NULL end) as week5s
, max(case when (week_num=5) then last_date else NULL end) as week5e
from week_dates(current_date)
) w ;
As before I have wrapped the above in a SQL function and provide an example here.
I would first simplify to:
extract(day from do_date)::int / 7 + 1 as week_in_month
then pivot on that using crosstab().
I want to count the number of months between two dates.
Doing :
SELECT TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP '2011-04-30 14:38:40';
Returns :
0 years 0 mons 409 days 20 hours 0 mins 0.00 secs
and so:
SELECT extract(month from TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP '2011-04-30 14:38:40');
returns 0.
age function returns interval:
age(timestamp1, timestamp2)
Then we try to extract year and month out of the interval and add them accordingly:
select extract(year from age(timestamp1, timestamp2)) * 12 +
extract(month from age(timestamp1, timestamp2))
Please note that the most voted answer by #ram and #angelin is not accurate when you are trying to get calendar month difference using.
select extract(year from age(timestamp1, timestamp2))*12 + extract(month from age(timestamp1, timestamp2))
for example, if you try to do:
select extract(year from age('2018-02-02'::date, '2018-03-01'::date))*12 + extract(month from age('2018-02-02'::date , '2018-03-01'::date))
the result will be 0 but in terms of months between March from February should be 1 no matter the days between dates.
so the formula should be like the following saying that we start with timestamp1 and timestamp2:
((year2 - year1)*12) - month1 + month2 = calendar months between two timestamps
in pg that would be translated to:
select ((extract('years' from '2018-03-01 00:00:00'::timestamp)::int - extract('years' from '2018-02-02 00:00:00'::timestamp)::int) * 12)
- extract('month' from '2018-02-02 00:00:00'::timestamp)::int + extract('month' from '2018-03-01 00:00:00'::timestamp)::int;
you can create a function like:
CREATE FUNCTION months_between (t_start timestamp, t_end timestamp)
RETURNS integer
AS $$
select ((extract('years' from $2)::int - extract('years' from $1)::int) * 12)
- extract('month' from $1)::int + extract('month' from $2)::int
$$
LANGUAGE SQL
IMMUTABLE
RETURNS NULL ON NULL INPUT;
The age function give a justified interval to work with:
SELECT age(TIMESTAMP '2012-06-13 10:38:40', TIMESTAMP '2011-04-30 14:38:40');
returns 1 year 1 mon 12 days 20:00:00, and with that you can easily use EXTRACT to count the number of months:
SELECT EXTRACT(YEAR FROM age) * 12 + EXTRACT(MONTH FROM age) AS months_between
FROM age(TIMESTAMP '2012-06-13 10:38:40', TIMESTAMP '2011-04-30 14:38:40') AS t(age);
If you will do this multiple times, you could define the following function:
CREATE FUNCTION months_between (t_start timestamp, t_end timestamp)
RETURNS integer
AS $$
SELECT
(
12 * extract('years' from a.i) + extract('months' from a.i)
)::integer
from (
values (justify_interval($2 - $1))
) as a (i)
$$
LANGUAGE SQL
IMMUTABLE
RETURNS NULL ON NULL INPUT;
so that you can then just
SELECT months_between('2015-01-01', now());
SELECT date_part ('year', f) * 12
+ date_part ('month', f)
FROM age ('2015-06-12', '2014-12-01') f
Result: 6 Months
Gives the differenece of months of two dates
SELECT ((extract( year FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract( year FROM TIMESTAMP '2011-04-30 14:38:40' )) *12) + extract(MONTH FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract(MONTH FROM TIMESTAMP '2011-04-30 14:38:40' );
The Result : 14
Have to extract months seperately for both the dates and then the difference of both the results
Here is a PostgreSQL function with the exact same behavior as the Oracle MONTHS_BETWEEN function.
It has been tested on a wide range of years (including leap ones) and more than 700k combinations of dates (including end of every months).
CREATE OR REPLACE FUNCTION months_between
( DATE,
DATE
)
RETURNS float
AS
$$
SELECT
(EXTRACT(YEAR FROM $1) - EXTRACT(YEAR FROM $2)) * 12
+ EXTRACT(MONTH FROM $1) - EXTRACT(MONTH FROM $2)
+ CASE
WHEN EXTRACT(DAY FROM $2) = EXTRACT(DAY FROM LAST_DAY($2))
AND EXTRACT(DAY FROM $1) = EXTRACT(DAY FROM LAST_DAY($1))
THEN
0
ELSE
(EXTRACT(DAY FROM $1) - EXTRACT(DAY FROM $2)) / 31
END
;
$$
LANGUAGE SQL
IMMUTABLE STRICT;
This function requires a LAST_DAY function (behaving the same as Oracle's one) :
CREATE OR REPLACE FUNCTION last_day
( DATE
)
RETURNS DATE
AS
$$
SELECT
(DATE_TRUNC('MONTH', $1) + INTERVAL '1 MONTH' - INTERVAL '1 DAY')::date
;
$$
LANGUAGE SQL
IMMUTABLE STRICT;
I had the same problem once upon a time and wrote this ... it's quite ugly:
postgres=> SELECT floor((extract(EPOCH FROM TIMESTAMP '2012-06-13 10:38:40' ) - extract(EPOCH FROM TIMESTAMP '2005-04-30 14:38:40' ))/30.43/24/3600);
floor
-------
85
(1 row)
In this solution "one month" is defined to be 30.43 days long, so it may give some unexpected results over shorter timespans.
Extract by year and months will floor on months:
select extract(year from age('2016-11-30'::timestamp, '2015-10-15'::timestamp)); --> 1
select extract(month from age('2016-11-30'::timestamp, '2015-10-15'::timestamp)); --> 1
--> Total 13 months
This approach maintains fractions of months (thanks to tobixen for the divisor)
select round(('2016-11-30'::date - '2015-10-15'::date)::numeric /30.43, 1); --> 13.5 months
Try this solution:
SELECT extract (MONTH FROM age('2014-03-03 00:00:00'::timestamp,
'2013-02-03 00:00:00'::timestamp)) + 12 * extract (YEAR FROM age('2014-03-03
00:00:00'::timestamp, '2013-02-03 00:00:00'::timestamp)) as age_in_month;
SELECT floor(extract(days from TIMESTAMP '2012-06-13 10:38:40' - TIMESTAMP
'2011-04-30 14:38:40')/30.43)::integer as months;
Gives an approximate value but avoids duplication of timestamps. This uses hint from tobixen's answer to divide by 30.43 in place of 30 to be less incorrect for long timespans while computing months.
I made a function like this:
/* similar to ORACLE's MONTHS_BETWEEN */
CREATE OR REPLACE FUNCTION ORACLE_MONTHS_BETWEEN(date_from DATE, date_to DATE)
RETURNS REAL LANGUAGE plpgsql
AS
$$
DECLARE age INTERVAL;
declare rtn real;
BEGIN
age := age(date_from, date_to);
rtn := date_part('year', age) * 12 + date_part('month', age) + date_part('day', age)/31::real;
return rtn;
END;
$$;
Oracle Example)
SELECT MONTHS_BETWEEN
(TO_DATE('2015-02-02','YYYY-MM-DD'), TO_DATE('2014-12-01','YYYY-MM-DD') )
"Months" FROM DUAL;
--result is: 2.03225806451612903225806451612903225806
My PostgreSQL function example)
select ORACLE_MONTHS_BETWEEN('2015-02-02'::date, '2014-12-01'::date) Months;
-- result is: 2.032258
From the result you can use CEIL()/FLOOR() for rounding.
select ceil(2.032258) --3
select floor(2.032258) --2
Try;
select extract(month from age('2012-06-13 10:38:40'::timestamp, '2011-04-30 14:38:40'::timestamp)) as my_months;
In PostgreSQL 8.4, given a date, if that date is not a Friday, I would like to find the date of the previous Friday. Can someone tell me if there is an inbuilt function or give the logic behind getting my own function.
Try this, works on other days too, blog about it http://www.ienablemuch.com/2010/12/finding-previous-day-of-week.html
create or replace function previous_date_of_day(the_date date, dow int) returns date
as
$$
select
case when extract(dow from $1) < $2 then
$1 - ( extract(dow from $1) + (7 - $2) )::int
else
$1 - ( extract(dow from $1) - $2)::int
end;
$$ language 'sql';
select to_char(z.ds, 'Mon dd yyyy dy') as source,
to_char( previous_date_of_day(z.ds, 5), 'Mon dd yyyy dy') as dest
from
(
select 'Dec 1 2010'::date + x.n as ds
from generate_series(0,17) as x(n)
) as z
You solve it without using case:
select
the_date
from
(
select
now()::date - num as the_date, -- generate rows of possible dates
extract(dow from (now()::date - num)) -- dow for the where condition
from (select generate_series(0,6) as num) as t
) as days
where date_part = 5;
SELECT
CASE
-- 1. if Friday, return date
WHEN EXTRACT(DOW FROM my_date) = 5
THEN my_date
-- 2. if Saturday, subtract 1
WHEN EXTRACT(DOW FROM my_date) = 6
THEN my_date - INTERVAL '1 day'
-- 3. all other days of the week, subtract `DOW + 2` from my_date
-- should be ELSE for future-proofing ;-) MB
ELSE -- WHEN EXTRACT(DOW FROM my_date) < 5 THEN
my_date - ((EXTRACT(DOW FROM my_date) + 2)::TEXT||'days')::INTERVAL
END AS tgif
FROM
my_table
WHERE
my_date IS NOT NULL
select case when extract(dow from your_date) < 5 then
your_date - (extract(dow from your_date) + integer '2')
else when extract(dow from your_date) > 5 then
your_date - integer '1'
else
your_date
end
Reference http://developer.postgresql.org/pgdocs/postgres/functions-datetime.html