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Split quarters to individual months
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I need to convert a yyyy/dd varchar type data( ex: 2021/03) to a monthly sort of date. (Ex: 2021/01, 2021/02, 2021/03). So, i need to convert the quarterly format to a monthly format in snowflake. Can we do this?
I tried many things but didn't get the expected results
select TO_DATE(date_column, 'YYYY/MM')
It may help to have a calendar table in your environment to deal with more complicated date conversions like this. An example of what this table may look like is in this CTE below. The SELECT statement following converts a string date of format YYYY/QQ of '2021/03' to YYYY/MM
WITH calendar AS
(
SELECT
dateadd('DAY', seq4(), '2000-01-01'::DATE) as calendar_date,
MONTH(calendar_date) as month_of_year,
QUARTER(calendar_date) as quarter_of_year,
YEAR(calendar_date) as year_of_calendar,
DAY(calendar_date) as day_of_month,
WEEK(calendar_date) as week_of_year
FROM table(generator(rowcount => 365*50))
)
SELECT DISTINCT '2021/03' as YYYYQQ_date,
year_of_calendar || '/' || LPAD(month_of_year, 2, '0') as YYYYMM
FROM calendar
WHERE year_of_calendar = strtok(YYYYQQ_date, '/', 1)
AND quarter_of_year = strtok(YYYYQQ_date, '/', 2);
2021/07
2021/08
2021/09
That's merely an example for turning yyyy/qq into multiple yyyy/mm outputs for that quarter, but this same logic can be applied to any date-part conversion and the calendar table can be customized to hold even organization-specific date things like oddball fiscal periods, company holiday flags, or a business day flag (as an example).
Related
I got the following SQLite database:
What I'm trying to do is: sum all values based on quantidade grouped by Month/Year (e.g.: 2022/08)
The result I'm expecting grouped by year/month:
My SQL code:
SELECT
data, SUM(quantidade) AS sum
FROM stock_tracking_Negociacao
WHERE mercado = 'Futuro'
GROUP BY strftime('%Y', data), strftime('%m', data)
Any help?
We can use SUBSTR() here to isolate the month and year, then aggregate:
SELECT SUBSTR(data, 4, 7) AS ym, SUM(quantidade) AS sum
FROM stock_tracking_Negociacao
WHERE mercado = 'Futuro'
GROUP BY 1;
Note that SQLite does not have a formal date type, but rather stores dates as strings.
For SQLite the only valid text-based date format is YYYY-mm-dd.
When you use any other format with date functions like strftime() the result is null and this is the main reason that your code does not work.
Since you mention that in the database the dates have the format YYYY/mm/dd you can update to the correct format:
UPDATE stock_tracking_Negociacao
SET data = REPLACE(data, '/', '-');
and then your query should be:
SELECT strftime('%Y-%m', data) AS year_month,
SUM(quantidade) AS sum
FROM stock_tracking_Negociacao
WHERE mercado = 'Futuro'
GROUP BY year_month;
I have an INT64 column called "Date" which contains many different numbers like: "20210209" or "20200305". I want to turn those numbers into a date with this format: MM-YYYY (so in these cases, 02-2021 and 03-2020). Ultimately I want to sum all the data in each month together. The problem is that BigQuery can't convert INT64 to date, only to strings. I'm not sure if I should convert to a string and then to a date or if there is a better way.
Although converting to a string then a date both works and is very concise, over large enough numbers of rows (which may be the case in Big Query) you may be better off using integer maths and using DATE(year, month, day)...
https://cloud.google.com/bigquery/docs/reference/standard-sql/date_functions#date
SELECT
DATE(
DIV( 20210209 , 10000), -- Which gives 2021
DIV(MOD(20210209, 10000), 100), -- Which gives 02
MOD(20210209, 100) -- Which gives 09
)
You can convert the value to a string and use parse_date():
select parse_date('%Y%m%d', cast(20210209 as string))
Another option
select date,
regexp_replace('' || date, r'(\d{4})(\d{2})(\d{2})', r'\2-\1') as MM_YYYY
from your_table
if applied to sample data in your question - output is
Yet another option
select date,
format_date('%m-%Y', parse_date('%Y%m%d', '' || date)) as MM_YYYY
from your_table
with same output
I want to format date in a column where date formats are mixed like d/mm/yy, d/m/yy, dd/mm/yyyy where i want ot format all values should be in one format like mm/dd/yyyy in sqlite database
SQLite does not support built-in date and/or time storage class. Instead, it leverages some built-in date and time functions to use other storage classes such as TEXT, REAL, or INTEGER for storing the date and time values.
use the TEXT storage class for storing SQLite date and time https://www.sqlitetutorial.net/sqlite-date/
This will convert month, day, year divided by slashes into year, month, day divided by dashes. For example 2/19/1921 into 1921-2-19. Just uses basic SQL with a subquery.
SELECT
surveydate
,Substr(dayyear, Instr(dayyear, '/') + 1, 999) || '-' || -- Year
month || '-' || -- Month
Substr(dayyear, 0, Instr(dayyear, '/')) -- Day
AS surveydate2
FROM (
SELECT
surveydate,
Substr(surveydate, 0, Instr(surveydate, '/')) AS month,
Substr(surveydate, Instr(surveydate, '/') + 1, 999) AS dayyear
FROM "input_locations"
)
In SQL. How to convert a column A from (YYYY-MM-DD) to (YYYYMM)? I want to show the dates in YYYYMM format instead of YYYY-MM-DD.
Data type is TIMESTAMP. Using Teradata Studio 15.10.10.
For Teradata either use
to_char(tscol, 'YYYYMM') -- varchar result
or
extract(year from tscol) * 100 + extract(month from tscol) -- integer result
In Teradata you can format dates pretty much at will. To get YYYYMM, you would use
select <your date> (format 'yyyymm') (char(6))
Your date column needs to be actual date for this, not a string.
There are 3 functions you'll need.
MONTH() function. Returns the MONTH for the date within a range of 1 to 12 ( January to December). It Returns 0 when MONTH part for the date is 0.
YEAR() function. Returns a 4 digit YEAR.
CONCAT() function is used to concatenate two or more strings together.
So here's an example of combining the 3 functions.
SELECT CONCAT(YEAR('1969-02-18'),MONTH('1969-02-18'))
or you can do it in one with
select DATE_FORMAT('1969-02-18','%Y%m')
So to answer your question if it is referring to column A, you can use
SELECT DATE_FORMAT(A,'%Y%m')
SQL Fiddle:
http://www.sqlfiddle.com/#!9/a6c585/48362
You can use DATEPART to get the year and month parts of the date, cast to a varchar, pad and the concaternate.
SELECT DATEPART(YEAR,GETDATE())
SELECT DATEPART(MONTH,GETDATE())
SELECT CAST(DATEPART(YEAR,GETDATE()) AS VARCHAR(4)) + RIGHT('00' + CAST(DATEPART(MONTH,GETDATE()) AS VARCHAR(2)),2)
I am using Hive, so the SQL syntax might be slightly different. How do I get the data from the previous month? For example, if today is 2015-04-30, I need the data from March in this format 201503? Thanks!
select
employee_id, hours,
previous_month_date--YYYYMM,
from
employees
where
previous_month_date = cast(FROM_UNIXTIME(UNIX_TIMESTAMP(),'yyyy-MM-dd') as int)
From experience, it's safer to use DATE_ADD(Today, -1-Day(Today)) to compute last-day-of-previous-month without having to worry about edge cases. From there you can do what you want e.g.
select
from_unixtime(unix_timestamp(), 'yyyy-MM-dd') as TODAY,
date_add(from_unixtime(unix_timestamp(), 'yyyy-MM-dd'), -1-cast(from_unixtime(unix_timestamp(), 'd') as int)) as LAST_DAY_PREV_MONTH,
substr(date_add(from_unixtime(unix_timestamp(), 'yyyy-MM-dd'), -1-cast(from_unixtime(unix_timestamp(), 'd') as int)), 1,7) as PREV_MONTH,
cast(substr(regexp_replace(date_add(from_unixtime(unix_timestamp(), 'yyyy-MM-dd'), -1-cast(from_unixtime(unix_timestamp(), 'd') as int)), '-',''), 1,6) as int) as PREV_MONTH_NUM
from WHATEVER limit 1
-- today last_day_prev_month prev_month prev_month_num
-- 2015-08-13 2015-07-30 2015-07 201507
See Hive documentation about date functions, string functions etc.
below works across year boundaries w/o complex calcs:
date_format(add_months(current_date, -1), 'yyyyMM') --previous month's yyyyMM
in general,
date_format(add_months(current_date, -n), 'yyyyMM') --previous n-th month's yyyyMM
use proper sign for needed direction (back/ahead)
You could do (year('2015-04-30')*100+month('2015-04-30'))-1 for the above mentioned date, it will return 201503 or something like (year(from_unixtime(unix_timestamp()))*100+month(from_unixtime(unix_timestamp())))-1 for today's previous month. Assuming your date column is in 'yyyy-mm-dd' format you can use the first example and substitute the date string with your table column name; for any other format the second example will do, add the column name in the unix_timestamp() operator.
Angelo's reply is a good start but it returns 201500 if the original date was 2015-01-XX. Building on his answer, I suggest using the following:
IF(month(${DATE}) = 1,
(year(${DATE})-1)*100 + 12,
year(${DATE})*100 + month(${DATE})-1
) as month_key
provided you get rid of those hyphens in your input string , previous date's month id in YYYYMM format you can get by:-
select if( ((${hiveconf:MonthId}-1)%100)=0 ,${hiveconf:MonthId}-89,${hiveconf:MonthId}-1 ) as PreviousMonthId;