I have 3 dropdown lists in a form for the user to input the date by day, month and year. How can I make that once t is the date that been input by the user, it will change the status from PUBLISH to UNPUBLISHED in the database?
Below is my codes:
Need your help. Thanks.
Why don't you use the DateSerial function to convert the separate values into date values, and use that to compare two dates.
From the page at W3Schools:
The DateSerial function returns a Variant of subtype Date for a specified year, month, and day.
Syntax
DateSerial(year,month,day)
Parameter Description
year Required. A number between 100 and 9999, or a numeric expression. Values between 0 and 99 are interpreted as the years 1900–1999. For all other year arguments, use a complete four-digit year
month Required. Any numeric expression
day Required. Any numeric expression
Examples
Example 1
document.write(DateSerial(2010,2,3))
The output of the code above will be:
2/3/2010
Example 2
Subtract 10 days:
document.write(DateSerial(2010,2,3-10))
The output of the code above will be:
1/24/2010
Related
I have a data written in specific expression. To simplify the data, here is the example I made:
df<-data.frame(date=c(2012034,2012044,2012051,2012063,2012074),
math=c(100,100,23,46,78))
2012034 means 4th week of march,2012. Likewise 2012044 means 4th week of April,2012. I was trying to make the values of date expressing some order. The reason why I have to do this is because when I don't change them to time expressions, x axis of the scatter plot looks really weird.
My goal is this:
Find the oldest date in date column and name it as 1. In this case, 2012034 should be 1. Next, find the second oldest date in date column and calculate how many weeks passed after that date. The second oldest date in date is 2012044.So, 5 weeks after the oldest date 2012034. So it should be changed as 1+5=6. So, likewise, I want to number the date to indicate how many weeks have passed since the oldest date
One way to do it is by also specifying the day of the week and subtract it at the end, i.e.
as.Date(paste0(df$date, '-1'), '%Y%m%U-%u') - 1
#[1] "2012-03-22" "2012-04-22" "2012-05-01" "2012-06-15" "2012-07-22"
In Stata I have a variable yearmonth which is formatted as 201201, 201202 etc. for the years 2012 - 2019, monthly with no gaps. When I format the variable as
format yearmonth %tm
The results look like: 2.0e+05 for all periods, with the exact same number each time. A Dickey-Fuller test tells me I have gaps in my data (I don't) and a tsfill command generates dozens of empty observations between each period.
How do I properly format my yearmonth variable so I can set it as a monthly date?
You do have gaps — between 201212 and 201301, for example. Consider a statement like
gen wanted = ym(floor(yearmonth/100), mod(yearmonth, 100))
which parses your integers like 201201 into year and month components. So floor(201201/100) is floor(2012.01) and so 2012 while mod(201201, 100) is 1. The two components are then the arguments of ym() which expects a year and a month argument.
Then and only then will your format statement do you want. That command won’t create date variables.
See help datetime in Stata for more information and Problem with displaying reformatted string into a four-digit year in Stata 17 for an explanation of the difference between a date value and a date display format.
I'm using ControlM and in a command, I would like to find a variable that gives me the date in this format : YYYYMM
I found there is %%$DATE variable but it gives YYYYMMDD
Thanks for you help
It is possible to define and concatenate a variable that will represent the date in such a format.
These are available:
Day DD, %%DAY,
Month MM, %%MONTH,
Year YY, %%YEAR,
Year YYYY, %%$YEAR
Prefer %%$OYEAR AND %%OMONTH over %%$YEAR and %%MONTH
I suggest using the variables %%$OYEAR and %%OMONTH over %%$YEAR and %%MONTH. The reason is that date variables beginning with O refer to processing dates and do not necessarily coincide with the system date. For this case you could use any of the following options:
1. YYYYMM = %%$OYEAR.%%OMONTH
2. YYYYMM = %%SUBSTRING %%$ODATE 1 6
The $ symbol preceding the variable %%$OYEAR or %%$ODATE indicates that the year is returned in 4-digit format, instead of OYEAR or ODATE which print the year with only 2 digits.
The dot (.) character is used for concatenate variables.
For example: For the order day May 29, 2020.
1. %%$ODATE would print 20200529
2. %%ODATE would print 200529
I tried to group my daily data by week (given a reference date) to generate a smaller panel data set.
I used postgres before and there it was quite easy:
CREATE TABLE videos_weekly AS SELECT channel_id,
CEIL(DATE_PART('day', observation_date - '2016-02-10')/7) AS week
FROM videos GROUP BY channel_id, week;
But it seems like it is not possible to subtract a timestamp with a date string in Drill. I found the AGE function, which returns an interval between two dates, but how to convert this into an integer (number of days or weeks)?
DATE_SUB may help you here. Following is an example:
SELECT extract(day from date_sub('2016-11-13', cast('2015-01-01' as timestamp)))/7 FROM (VALUES(1));
This will return number of weeks between 2015-01-01 and 2016-11-13.
Click here for documentation
I have two columns in my table one is having a date in yyyymm format and other column has some integer values between 1 to 50. How can I add these two fields and get a date value?
For example: 201402 + 12 should give me 201502 as an answer!
I assume you don't really have a DATE column but a varchar column that stores a month specification in the format yyyymm.
If you want to make use of Oracle's date arithmetic you first need to convert this "month" into a real date.
Something like this:
select to_char(add_months(to_date('201402', 'yyyymm'), 12), 'yyyymm')
from dual;
You will need to replace the character literal '201402' with a reference to your column.