Convert parameter in Select statement - stored-procedures

I am trying to convert a date parameter in my select statement but i wanted to format it and currently it is not accepting how i have it written -- Any advice would be appreciated
SELECT #cases = COALESCE(#cases+ #linebreak + C.CaseNumber, C.CaseNumber)+"-"+CT.[Description]+"-"+CONVERT(VARCHAR,#assignment,100)AS A.DateOn
This is how i had it before the formatting and it was converting and working fine:
SELECT #cases = COALESCE(#cases+ #linebreak + C.CaseNumber, C.CaseNumber)+"-"+CT.[Description]+"-"+CONVERT(VARCHAR, A.DateOn)
Not sure if more information will be needed or not but let me know.

Guessing, your last bit should be
...+CONVERT(VARCHAR,A.DateOn,101)
Where does #assignment come from? Your unformatted code doesn't mention it...
Note (after question comment) 4/14/2010 is harder to do than 04/14/2010. You have to evaluate day, month and year separately to remove leading zeros

Related

How to generate sequences in Google Sheets

I'l like to create a sequence of dates using "sequence formula" with a start and end date. I mean, something like this:
start date = 2020-01-01
end date = 2022-12-01
I just also want to bring the first day of the month, like the following example:
2020-01-01
2020-02-01
2020-03-01
......
2022-10-01
2022-11-01
2022-12-01
How can I get this? Is there another way to do it?
Thanks!!!
There are several ways to approach this. Here is one I recommend:
=ArrayFormula(DATE(2020,SEQUENCE(DATEDIF("2020-01-01", "2022-12-01", "m")+1),1))
You'll replace the first 2020 with the year of your starting date and the next two string-dates with your full start and end date. That's it.
Just make sure you leave enough cells open below this formula to accommodate the number of dates you're requiring the formula to output.
Format the output range in the date format you prefer (Format > Number...).
Simple SEQUENCE() should work. Try-
=SEQUENCE((B1-A1)+1,1,A1)

Set new year for date column in rails query

I have the contract start of a number of companies, and I want to report on each contract year by creating a column with the contract start updated to a select year. There are a number of solutions in SQL involving functions like DATE_ADD or DATEFROMPARTS, but I'm having trouble adapting it to rails (if those functions are available at all).
The closest I've gotten is: Company.select("contract_start + '1 YEAR'::INTERVAL as new_contract_start"). This adds 1 year to each contract start but doesn't take into account contracts older than a year (or started the same year). I've also tried the following but again run into syntax errors:
new_year = 2020
Company.select("contract_start + '#{new_year} - EXTRACT (YEAR from contract_start) YEAR'::INTERVAL")
I'm looking for a solution that can either:
Directly set the year to what I want
Add a variable amount of years based on its distance from the desired year
I'm on Ruby 2.3.3
I think the key here was finding functions compatible with the PostgreSQL that my database was built on. Once I started searching for the functions I thought would help and their PostgreSQL equivalents, I found more compatible solutions, such as: NUMTODSINTERVAL in PostgreSQL
I ended up with:
contract_start_year = 2020
Company.select("contract_start + make_interval(years => CAST (#{contract_start_year} - EXTRACT (YEAR from contract_start) as INT))
I've also made it a bit smarter by adding the number of years required to get the latest contract date without going over the report date. This would be problematic if the report start date was "2020-01-01" but the contract start was "2017-06-01". Setting the contract date to "2020-06-01" would overshoot the intentions of the report.
report_start = "`2020-07-01`"
Company.select("contract_start + make_interval(years => CAST (EXTRACT (YEAR FROM AGE(CAST (#{start_quotations} AS DATE), contract_start)) AS INT)) as new_contract_year")
Note the additional single quotes in report_start since the SQL code need to read a string to convert it to a date
There might be other methods that can "build" the date directly, but this methods works well enough for now.

Generating a random date without time in Ruby

I have a date of birth column on my user table that takes a DATE. As this datatype appears as YYYY-MM-DD, I assume that when inputting a date to the database it must have the format, for example: 2013-12-26.
I have seen methods on StackOverflow for creating a random DateTime in Ruby, such as here. However, after much searching I can't find a way to generate a random date without the time, for example in the past 100 years, and have it properly formatted for the DATE datatype. In Rails, what is the best way to generate a random date without the time?
This seems to work:
def rand_date(days)
date = Date.today-rand(days)
date.to_s(:db)
end
But is there a more elegant solution that comes with Rails? I am new to Rails and programming, so any assistance would be most helpful!
Your method is correct. If you are using Rails, there are some trivial improvements such as
def rand_date(days)
rand(days).days.ago(Date.today)
end
which is mostly equivalent to
def rand_date(days)
rand(days).days.ago.to_date
end
The second version is less efficient because it creates more Date/Time objects during the internal conversions.
Apply to_s(:db) if you need the Date to be formatted as String.
A different approach would require you to construct a date passing the result of a rand to Date.new.
This is in core ruby:
1 #!usr/bin/ruby
2
3 require 'date'.
4
5 10.times do |t|
6 random_year = Random.new.rand(2000..2014) # custom range for years
7 random_month =Random.new.rand(1..12)
8 random_day = Random.new.rand(1..30)
9 puts "#{Date.new(random_year,random_month,random_day)}"
10 end
2014-11-29
2010-10-20
2006-02-23
2009-09-17
2006-01-14
2009-01-06
2002-07-06
2005-11-05
2013-06-20
2005-12-02
Here is something I used to generate random birth dates when populating a customer database. It works on days, and in this example, gives random dates between 1967-01-09 and 1993-01-12 by using the Date#jd method:
Date.jd(2439500 + rand(9500))
You can twiddle the dates generated by setting the base (in this case 2439500, which is 1967-01-09) and the random number to increase or decrease the range of dates produced.
Example:
irb(main):043:0> 10.times { puts Date.jd(2439500 + rand(9500)) }
1973-06-07
1973-11-09
1983-07-27
1990-11-03
1967-06-18
1967-06-20
1970-07-28
1990-05-13
1986-11-26
1989-02-15

Good design for incrementing / decrementing week view of a to do list?

I am looking for advice on how to design a web app that shows a week view for a todo list.
The current layout is a table with days of the week as columns with items to be done listed below. I have gotten something functional by calculating the header dates in the controller with:
#week_starting = Date.today.at_beginning_of_week if #week_starting.blank?
Other weekday headers are shown by incrementing #week_starting.
I have created individual instance variables such as #to_do_on_mon, #to_do_on_tue, etc. to show what is to be done that day (and then iterate through them to list out the items).
Here is an example of one:
#to_do_on_mon = BatchTicket.where("Date(load_date) =?", #week_starting.strftime
How would you improve on this design to easily increment/decrement the week shown and list out items to be done?
First of all, just a personal preference, but I think it would make the code nicer changing the first line to this:
#week_starting ||= Date.today.at_beginning_of_week
It's shorter and it's clearer.
I'm not sure I understand what you're trying to do with the Date(load_date) there, but since you have ActiveSupport available, I would probably do something more readable like this:
#to_do_on_mon = BatchTicket.where date: (#week_starting + 0.days)
#to_do_on_tue = BatchTicket.where date: (#week_starting + 1.days)
# etc.
Or, depending on what your page/app looks like, you can do it in one query if you need to:
BatchTicket.where('your date range').group_by { |b| b.date.strftime("%u") }
Will return an array grouped into days (1..7).
Hope that helps

I'm getting "Invalid month in date" trying to run this?

I'm trying to run the following db command against Informix:
delete from table1
where u_id in (select u_id
from table2
where c_id in (select c_id
from ptable
where name = 'Smith'
and dob = '29-08-1946'));
I pass this in as a string to the db.ExecuteNonQuery method in the MS Data Application block and I get the above error?
To get the date format '29-08-1946' to work, you need your DBDATE environment variable set to a value such as "DMY4-" (or "DMY4/"). These are standard variations for the UK (I used them for years; I now use "Y4MD-" exclusively, which matches both ISO 8601:2004 (Date formats) and ISO 9075 (SQL), except when debugging someone else's environment). There are other environment variables that can affect date formatting - quite a lot of them, in fact - but DBDATE takes priority over the others, so it is the big sledgehammer that fixes the problem.
One of the problems is that your notation using a plain string is not portable between US and UK (and ISO) settings of DBDATE. If you have a choice, the neutral constructor for dates is the MDY() function:
WHERE dob = MDY(8,29,1946)
This works regardless of the setting of DBDATE. You can probably use TO_DATE() too:
SELECT TO_DATE('29-08-1946', '%d-%m-%Y') FROM dual;
This generated '1946-08-29 00:00:00.00000' for me - the function generates a DATETIME YEAR TO FRACTION(5) value, but those convert reliably to DATE values in Informix.
You can also use the DATE() function or an explicit cast to DATE (either CAST('29-08-1946' AS DATE) or '29-08-1946'::DATE), but both of those are subject to the whims of the locale of the users.
Your date field is improperly formatted. Since there is no 29th month in the year 1946 that is what is causing the error.
I'd try just swapping the month and day. 08-29-1946.
The way the day and month parts of a date string are read in can depend on your computer's culture settings.
It is always safer to pass date strings to a database in the form 'dd-MMM-yyyy' (i.e. '29-aug-1946')
It's even safer to pass them as YYYY-MM-DD, the dd-MMM-yyyy in that example will fail on a server with a (for example) French locale.

Resources