I am trying to generate a string array of weekdays and use it find how many times each day appears in a month
I am using standard sql on BigQuery
My query would look like
with weeks as (select array['SUNDAY','MONDAY','TUESDAY','WEDNESDAY','THURSDAY','FRIDAY','SATURDAY'] as wk)
select DATE_DIFF('2019-01-31','2019-01-01',WEEK(wk)) AS week_weekday_diff
from weeks, unnest(wk) as wk
The query however fails with the error A valid date part argument for WEEK is required, but found wk. wk is a column value having the Days of Week, WEEK is a Functions which expects a literal DAYOFWEEK. Is there a way i pass the column value as arguments
Below is for BigQuery Standard SQL
error "A valid date part argument for WEEK is required, but found wk"
WEEK(<WEEKDAY>): Valid values for WEEKDAY are literal SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, and SATURDAY.
... Is there a way i pass the column value as arguments?
If you wish - you can submit feature request at https://issuetracker.google.com/issues/new?component=187149&template=0
find how many times each day appears in a month
To get your expected result and overcome above "issue" you can approach task from opposite angle - just extract weekdays positions and then do needed stats as in example below
#standardSQL
WITH weekdays AS (SELECT ['SUNDAY','MONDAY','TUESDAY','WEDNESDAY','THURSDAY','FRIDAY','SATURDAY'] AS wk)
SELECT wk[ORDINAL(pos)] weekday, COUNT(1) cnt
FROM weekdays,
UNNEST(GENERATE_DATE_ARRAY('2019-01-01','2019-01-31')) day,
UNNEST([EXTRACT(DAYOFWEEK FROM day)]) pos
GROUP BY pos, weekday
ORDER BY pos
with result
Row weekday cnt
1 SUNDAY 4
2 MONDAY 4
3 TUESDAY 5
4 WEDNESDAY 5
5 THURSDAY 5
6 FRIDAY 4
7 SATURDAY 4
Trying your query, what I have noticed to be returning an error is:
select DATE_DIFF('2019-01-31','2019-01-01',WEEK('WEDNESDAY')) AS week_weekday_diff;
as the function WEEK(< WEEKDAY >) is expecting something like:
select DATE_DIFF('2019-01-31','2019-01-01',WEEK(`WEDNESDAY`)) AS week_weekday_diff;
OR
select DATE_DIFF('2019-01-31','2019-01-01',WEEK(WEDNESDAY)) AS week_weekday_diff;
I think that the WEEK(< WEEKDAY >) only accepts the weekdays in the format exposed here, so no strings should be valid.
Related
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
I need some help in writing an SQL in SQL Server where I need to count number of rows group by weeks. There is a tricky description of week which is following
- For any date before 08/13/2015 the week is of 7 days (i.e. from Thu through Wed)
- For date 08/13/2015 the week is consider a 9 day week (i.e. from Thursday through Friday so its between 08/13/2015 through 08/21/2015)
- For date 08/22/2015 the week is back to 7 days (i.e. Sat through Friday)
Now having said all the above the result I want to see in my report is the following way . NOTE: WE column in the below attached image is the last day of the week for the range.
Sample Result Image
Just write a case statement for the 3 different options. You can find the start day with something like this:
DATEADD(week, DATEDIFF(day, 3,getdate()) / 7, 3) -- Thursdays
DATEADD(week, DATEDIFF(day, 5,getdate()) / 7, 5) -- Saturdays
The numbers 3 and 5 come from the fact that day 0 (=1.1.1900) is Monday.
If you use this a lot, it might be a good idea to write a inline table valued function to return the dates you need.
I'm looking for a workaround or hive date functions that gives day of the week ,
Sunday - 1
Monday - 2
Tuesday - 3
Wednesday - 4
Thursday - 5
Friday - 6
Saturday - 7
Requirement in detail : I'm looking for a function that takes date string (YYYYMMDD) as input and outputs the day of the week as per the above table.
Consider using from_unixtime(your date,'u') - this will return day number of week starting from Monday=1.
If your date is not in unixtime format, you can use the following instead:
from_unixtime(unix_timestamp('20140112','yyyyMMdd'),'u')
see: http://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html for simple date format documentation.
You can now use date_format (Hive 1.2):
hive> select date_format('2016-12-01' ,'u');
OK
4
select pmod(datediff(your_date,'1900-01-07'),7) + 1 as WeekDay from your_table
arbitrary start date picked (1900-01-07)
calculates the mod 7 day of week (plus 1 to start at 1 instead of zero)
Expanding on iggy's answer, here is the query to get the days of the week. Adjust the query to set the first day of the week as necessary.
SELECT current_date AS `Date`,
CASE date_format(current_date,'u')
WHEN 1 THEN 'Mon'
WHEN 2 THEN 'Tues'
WHEN 3 THEN 'Wed'
WHEN 4 THEN 'Thu'
WHEN 5 THEN 'Fri'
WHEN 6 THEN 'Sat'
WHEN 7 THEN 'Sun'
END AS day_of_week
From Hive 2.2 there is another possibility:
hive> select extract(dayofweek FROM your_date) FROM your_table;
As I said you need to write a UDF which will accept a string as parameter and return a string.
Inside the UDF you need to do these steps:
1.) Parse the input string using SimpleDateFormat(YYYYMMDD)
2.) Use the Below code to get the day of week:
Calendar c = Calendar.getInstance();
c.setTime(yourDate);
int dayOfWeek = c.get(Calendar.DAY_OF_WEEK);
3.) Use this dayOfWeek value in a case statement to get your weekday String and return that string.
Hope this helps...!!!
In my table I have for each day:
Id_______Startdate _______ EndDate __________MondayMorning _____MondayEvening ___TuMorning ....
121 _____2012-01-01________2012-12-31 ________2 ___________________2______________2
122 _____2012-02-01________2012-08-05 ________1 ___________________2______________3
I already generated a list of dates using the Startdate and EndDate.
I want to know if this dates belongs to a odd week or a even week so that I can filter days from my output that have number 3 Or 1 (see second record).
how can I filter the days that belongs to odd weeks and even weeks that have number 1 and 3?
You can use datepart with the wk argument to determine the week number:
SELECT datepart(wk, YourDate)
From there, you can use modulus to determine if the week number is even or odd:
SELECT datepart(wk, YourDate) % 2
This will return 0 for even numbered weeks, and 1 for odd numbered weeks.
Am trying to return the sum of each day of a week in mysql but it returns nothing despite having values for the 3rd Week of March 2010
SELECT SUM(expense_details_amount) AS total
FROM expense_details
WHERE YEAR(expense_details_date) = '2010'
AND MONTH(expense_details_date) = '03'
AND WEEK(expense_details_date) = '3'
GROUP BY DAY(expense_details_date)
How do I go about this?
According to the MySQL docs for WEEK:
This function returns the week number for date. The two-argument form of WEEK() allows you to specify whether the week starts on Sunday or Monday and whether the return value should be in the range from 0 to 53 or from 1 to 53. If the mode argument is omitted, the value of the default_week_format system variable is used. See Section 5.1.4, “Server System Variables”.
Their examples are:
mysql> SELECT WEEK('2008-02-20');
-> 7
mysql> SELECT WEEK('2008-02-20',0);
-> 7
mysql> SELECT WEEK('2008-02-20',1);
-> 8
mysql> SELECT WEEK('2008-12-31',1);
-> 53
So you should not be checking for week 3, but whatever week of the year the 3rd week in march is.
It also looks like these functions return integers instead of strings, so you might want to lose the single-quotes in your query.
Your where clause excludes all of the data, as the third week of the year is in January and you have also specified the month to be March. Try using a where clause of the form:
WHERE expense_details_date BETWEEN '2010-03-15' AND '2010-03-22'
WEEK() actually works like WEEKOFYEAR(), not WEEKOFMONTH(), so it gives values 0 through 53 as the result, by default.
If you're saying that the first week of the month is the first 7 days, not starting at the first Sunday, then you can use DAYOFMONTH(), and BETWEEN to get your third week:
SELECT SUM(expense_details_amount) AS total
FROM expense_details
WHERE YEAR(expense_details_date) = 2010
AND MONTH(expense_details_date) = 3
AND DAYOFMONTH(expense_details_date) BETWEEN 15 AND 21
GROUP BY DAYOFMONTH(expense_details_date)
Notice that I also changed the constants to numerals instead of Strings so MySQL doesn't have to do the conversion.