Extract day of week, day, month in string from JSON - iphone

I have a problem with the formatting of dates and I gently ask your help
I have a set of data in Json format including a date in the format listed below, example:
DateEvent ":" 02/24/2012 00:00:00 "
From this date I extract format string, locating the language, the following values​​:
Day of week: Friday (Italian)
Day: 24
Month: January (Italian)
Can anyone help me?

Use a NSDateFormatter to translate the string to a NSDate object. Then use NSDateComponents with that date to get the values you want.

Related

how to get a week's date with autohotkey

i'm using autohotkey to work with date, i need to catch the day a week ago
example
if today is the 28th then I have to take the 21st of last week
calendar
in the following script I take the current date
FormatTime, date, , dd/MM/yyyy
MsgBox %date%
I even thought of a logic, to take the current day subtract by 7 that will take the day a week ago. I need help to create a better script
28 - 7 = 21
if anyone can help me thanks :)
Just subtracting numbers would be bad when you encountered a change between months.
Would have to create custom logic for that.
Luckily AutoHotkey's += operator supports date/time math.
So this is all you need:
;we're starting off the date1 variable as blank,
;which means the current time will be used.
date1 += -7, days
FormatTime, finalDate, % date1, dd/MM/yyyy ;format the result to our desired format
MsgBox, % finalDate
I did it that way
FormatTime, date_, , dd
sub += date_-7
FormatTime date, , /MM/yyyy
MsgBox,%sub%%date%

How to convert a string date format `October 18th 2019` into a valid date `2019-10-17T23:00:00.000Z`

I need to convert a string date format like this October 18th 2019 ('MMMM Do YYYY') into a valid date 2019-10-17T23:00:00.000Z or similar 17/10/2019
i have tried using parsing the string into moment but i keep getting errors
update: I used moment('October 18th 2019').format(). received invalid date as the error, sorry I should clarify i'm trying to convert the string October 18th 2019 into a valid date format,
You simply need to supply the format string for the input (MMMM Do YYYY) when constructing the Moment object, with one of the following approaches:
// this way interprets the input at the start of the day in the local time zone
moment('October 18th 2019', 'MMMM Do YYYY')
// this way interprets the input at the start of the day in UTC
moment.utc('October 18th 2019', 'MMMM Do YYYY')
// this way interprets the input at the start of the day in a specific named time zone
// (requires the moment-timezone add-on)
moment.tz('October 18th 2019', 'MMMM Do YYYY', 'Europe/London')
Then you can format and/or convert it however desired. For example:
// this way keeps the local time, includes the local time offset when formatting
moment('October 18th 2019', 'MMMM Do YYYY').format()
// this way converts from local to utc before formatting
moment('October 18th 2019', 'MMMM Do YYYY').utc().format()
// this way converts from local to utc before formatting and includes milliseconds
moment('October 18th 2019', 'MMMM Do YYYY').toISOString()

How to print current Time and Date in Q Basic?

What is the command for displaying the time and date in qbasic? Could the syntax for the commands be given as well? And an explanation if possible?
You can use DATE$ and TIME$
These can also set the date and time as well.
The command for printing the time(current system time) is time$
The time$ is actually a function, in this case, no parameter is needed.
And the code is...
PRINT TIME$
The time is printed in hh: mm: ss format(hour: minutes: seconds).
And therefore the output would be something like this:
14:55:28
For printing the current system date, we use date$ function which is also a string function
The code is:
PRINT DATE$
The date is printed in mm-dd-yyyy format or month-day-year(American date format).
Hence the output will be:
02-17-2018
Hope it helps...
The QB date/time functions are:
DATE$ returns the date in a string in the form MM-DD-YYYY
TIME$ returns the time in a string in the form HH:MM:SS
When used as a command the date$ and time$ can be assigned to set the system date and time, for example DATE$ = "12-10-1990" or TIME$ = "12:10:10"
If the year is a leap year then the 29th day of February could be set. Otherwise if it is not a leap year then a syntax error will occur trying to set the date in February to the 29th.

Transform string monthly dates in Stata

I have a problem in Stata with the format of the dates. I believe it is a very simple question but I can't see how to fix it.
I have a csv file (file.csv) that looks like
v1 v2
01/01/2000 1.1
01/02/2000 1.2
01/03/2000 1.3
...
01/12/2000 1.12
01/02/2001 1.1
...
01/12/2001 1.12
The form of v1 is dd/mm/yyyy.
I import the file in Stata using import delimited ...file.csv
v1 is a string variable, v2 is a float.
I want to transform v1 in a monthly date that Stata can read.
My attempts:
1)
gen Time = date(v1, "DMY")
format Time %tm
which gives me
Time
3177m7
3180m2
3182m7
...
that looks wrong.
2) In alternative
gen v1_1=v1
replace v1_1 = substr(v1_1,4,length(v1_1))
gen Time_1 = date(v1_1, "MY")
format Time_1 %tm
which gives exactly the same result.
And if I type
tsset Time, format(%tm)
it tells me that there are gaps but there are no gaps in the data.
Could you help me to understand what I'm doing wrong?
Stata has wonderful documentation on dates and times, which you should read from beginning to end if you plan on using time-related variables. Reading this documentation will not only solve your current problem, but will potentially prevent costly errors in the future. The section related to your question is titled "SIF-to-SIF conversion." SIF means "Stata internal form."
To explain your current issue:
Stata stores dates as numbers; you interpret them as "dates" when you assign a format. Consider the following:
set obs 1
gen dt = date("01/01/2003", "DMY")
list dt
// 15706
So that date is assigned the value 15706. Let's format it to look like a day:
format dt %td
list
// 01jan2003
Now let's format it to be a month:
format dt %tm
list
// 3268m11
Notice that dt is just a number that you can format and use like a day or month. To get a "month number" from a "day number", do the following:
gen mt = mofd(dt) // mofd = month of day
format mt %tm
list
// dt mt
// 3268m11 2003m1
The variable mt now equals 516. January 2003 is 516 months from January 1960. Stata's "epoch time" is January 1, 1960 00:00:00.000. Date variables are stored as days since the epoch time, and datetime variables are stored as miliseconds since the epoch time. A month variable can be stored as months since the epoch time (that's how the %tm formatting determines which month to show).

How do I parse "YYYY-MM-DD" with joda time

I'm trying to use joda-time to parse a date string of the form YYYY-MM-DD. I have test code like this:
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("YYYY-MM-DD");
DateTime dateTime = dateDecoder.parseDateTime("2005-07-30");
System.out.println(dateTime);
Which outputs:
2005-01-30T00:00:00.000Z
As you can see, the DateTime object produced is 30 Jan 2005, instead of 30 July 2005.
Appreciate any help. I just assumed this would work because it's one of the date formats listed here.
The confusion is with what the ISO format actually is. YYYY-MM-DD is not the ISO format, the actual resulting date is.
So 2005-07-30 is in ISO-8601 format, and the spec uses YYYY-MM-DD to describe the format. There is no connection between the use of YYYY-MM-DD as a pattern in the spec and any piece of code. The only constraint the spec places is that the result consists of a 4 digit year folowed by a dash followed by a 2 digit month followed by a dash followed by a two digit day-of-month.
As such, the spec could have used $year4-$month2-$day2, which would equally well define the output format.
You will need to search and replace any input pattern to convert "Y" to "y" and "D" to "d".
I've also added some enhanced documentation of formatting.
You're answer is in the docs: http://www.joda.org/joda-time/apidocs/org/joda/time/format/DateTimeFormat.html
The string format should be something like: "yyyy-MM-dd".
The date format described in the w3 document and JodaTime's DateTimeFormat are different.
More specifically, in DateTimeFormat, the pattern DD is for Day in year, so the value for DD of 30 is the 30th day in the year, ie. January 30th. As the formatter is reading your date String, it sets the month to 07. When it reads the day of year, it will overwrite that with 01 for January.
You need to use the pattern strings expected by DateTimeFormat, not the ones expected by the w3 dat and time formats. In this case, that would be
DateTimeFormatter dateDecoder = DateTimeFormat.forPattern("yyyy-MM-dd");