I am creating a date dimension table using pentaho and im following the tutorial linked https://www.fabguy.de/howto-create-a-date-dimension-with-pdi-kettle.html
but when I preview my data, I found a mismatch between the column day of week and the column week_name : the week_name showed 'Monday' but the day of week showed '2' when the date is 1900-01-01.
The day of week was generated from the calculation 'Day of year of date A' in the calculator, and the week_name was generated from 'create a copy of field A'.
I noticed that the week_name is the correct one, but I really don't understand why the day of week is wrong.
the whole process looks like this:
enter image description here
the calculator named Calculate additional Dates looks like this:
enter image description here
the mismatch looks like this:
enter image description here
I really want to know how to make the day of week become the correct one, if anyone know how to fix it or the reason of it, plz leave a message here!
Thank you very much!!!
By default, Pentaho starts with Sunday (1) if you want to start with Monday try to use the Formula step with the WEEKDAY function instead.
The Formula step has 3 type
Type 1: Sunday is the first day of the week, with value 1; Saturday has value 7
Type 2: Monday is the first day of the week, with value 1; Sunday has value 7
Type 3: Monday is the first day of the week, with value 0; Sunday has value 6
For example, if today is Thursday I will get the following results
Related
When the client select start date ,end date and group by : weekly is there any way to force to show in the week column as first day of the week. For example if I choose 10/1/2019 as start date and 10/2/2019 as end date and group by : weekly then the week column should show 9/29/2019
Right now the expression I am using is
= Switch(Parameters!GroupbyTime=“4”, CDate(Fields!groupweek.value).ToString(“Mm/Dd/yyyy”)
But it’s giving error can somebody help me ??
You can subtract the number of the weekday from the date plus one day for the first day of the week.
=DATEADD("D", 1 - WEEKDAY(Parameters!START.Value), Parameters!START.Value)
I have an sheet with the Table, having Column A in Calendar Week. I would like to generate an chart with x axis as month with the data available in the table.
Could anyone help me to code it ? I am struck how I can convert an weeknumber in the column to month.
Could anyone suggest, how I can generate a chart for these data with month in my x/axis. I am not getting an clue to do this. Any lead would be helpful.
This is a formula to convert the week number in cell A1 into month
=MONTH(DATE(2017,1,-2)-WEEKDAY(DATE(2017,1,3))+A1*7)
Note that you also need to know the year the week number refers to, because it can differ from year to year. In this example it is fixed to 2017.
Note that week number != week number.
There are different methods how week numbers are calculated.
This formula example is based on ISO week numbers, with a week starting on Monday and the week containing the 1st Thursday of the year is considered week 1. For example, in the year 2016, the first Thursday is January 7, and that is why week 1 begins on 4-Jan-2016.
Here is some info about week number calculation based on different week numbers: Week Numbers In Excel
You need to use the formula
=TEXT((((A1-1)*7)+DATE(2017,1,1)),"m")
This will turn the week number into a date, then extract the month number
I have 70.000 rows of data, including a date time column (YYYY-MM-DD HH24-MM-SS.).
I want to split this data into 3 separate columns; Hour, day and Week number.
The date time column name is 'REGISTRATIONDATE' from the table 'CONTRACTS'.
This is what I have so far for the day and hour columns:
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour"
FROM CONTRACTS;
I have seen the options to get a week number for specific dates, this assignment concerns 70.000 dates so this is not an option.
You (the OP) still have to explain what week number to assign to the first few days in a year, until the first Monday of the year. Do you assign a week number for the prior calendar year? In a Comment I asked about January 1, 2017, as an example; that was a Sunday. The week from January 2 to January 8 of 2017 is "week 1" according to your definition; what week number do you assign to Sunday, January 1, 2017?
The straightforward calculation below assigns to it week number 0. Other than that, the computation is trivial.
Notes: To find the Monday of the week for any given date dt, we can use trunc(dt, 'iw'). iw stands for ISO Week, standard week which starts on Monday and ends on Sunday.
Then: To find the first Monday of the year, we can start with the date January 7 and ask for the Monday of the week in which January 7 falls. (I won't explain that one - it's easy logic and it has nothing to do with programming.)
To input a fixed date, the best way is with the date literal syntax: date '2017-01-07' for January 7. Please check the Oracle documentation for "date literals" if you are not familiar with it.
So: to find the week number for any date dt, compute
1 + ( trunc(dt, 'iw') - trunc(date '2017-01-07', 'iw') ) / 7
This formula finds the Monday of the ISO Week of dt and subtracts the first Monday of the year - using Oracle date arithmetic, where the difference between two dates is the number of days between them. So to find the number of weeks we divide by 7; and to have the first Monday be assigned the number 1, instead of 0, we need to add 1 to the result of dividing by 7.
The other issue you will have to address is to convert your strings into dates. The best solution would be to fix the data model itself (change the data type of the column so that it is DATE instead of VARCHAR2); then all the bits of data you need could be extracted more easily, you would make sure you don't have dates like '2017-02-29 12:30:00' in your data (currently, if you do, you will have a very hard time making any date calculations work), queries will be a lot faster, etc. Anyway, that's an entirely different issue so I'll leave it out of this discussion.
Assuming your REGISTRATIONDATE if formatted as 'MM/DD/YYYY'
the simples (and the faster ) query is based ond to to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW')
(otherwise convert you column in a proper date and perform the conversio to week number)
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour",
to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW') as "Week"
FROM CONTRACTS;
This is messy, but it looks like it works:
to_char(
to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS') +
to_number(to_char(trunc(to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS'),'YEAR'),'D'))
- 2,
'WW')
On the outside you have the solution previous given by others but using the correct date format. In the middle there is an adjustment of a certain number of days to adjust for where the 1st Jan falls. The trunc part gets the first of Jan from the date, the 'D' gets the weekday of 1st Jan. Since 1 represents Sunday, we have to use -2 to get what we need.
EDIT: I may delete this answer later, but it looks to me that the one from #mathguy is the best. See also the comments on that answer for how to extend to a general solution.
But first you need to:
Decide what to do dates in Jan before the first Monday, and
Resolve the underlying problems in the date which prevent it being converted to dates.
On point 1, if assigning week 0 is not acceptable (you want week 52/53) it gets a bit more complicated, but we'll still be able to help.
As I see it, on point 2, either there is something systematically wrong (perhaps they are timestamps and include fractions of a second) or there are isolated cases of invalid data.
Either the length, or the format, or the specific values don't compute. The error message you got suggests that at least some of the data is "too long", and the code in my comment should help you locate that.
Please help me with an Informx SQL query to be run every Friday.
Period: 1st of the Month to the Thursday before the Friday that it is being run on. I can't just choose date ranges as it will be an auto report so the dates needs to update automatically. Is there a way to do this?
You've got two issues: the start of the month and the date of last Thursday. The simplest expression to identify the 1st day of the current month is:
MDY(MONTH(TODAY), 1, YEAR(TODAY))
The way to identify the most recent Thursday is via a CASE statement on WEEKDAY(TODAY). Something like:
CASE
WHEN WEEKDAY(TODAY) < 5 -- (Sunday (0) - Thursday (4))
THEN TODAY - WEEKDAY(TODAY) - 3 -- Calc last Sunday, back 3 days
ELSE TODAY - WEEKDAY(TODAY) + 4 -- Calc last Sunday, forward 4 days
END
Note, you will still run into issues where the 1st of current month is later than the most recent Thursday. But hopefully the examples above will point you in the right direction.
Of course, if you know this report is only ever going to be run on Fridays, then calculating the most recent Thursday is just TODAY -1.
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).