Utilizing the DATE_SUBTRACT function in AQL, when handling dates near the end of the month, Arango seems to subtract 30 days instead of returning the actual previous month. e.g.:
for mo in 0..11
let month = date_subtract(date_now(),count,"month")
return month
returns
[
"2016-08-31T20:30:24.440Z",
"2016-07-31T20:30:24.441Z",
"2016-07-01T20:30:24.441Z",
"2016-05-31T20:30:24.441Z",
"2016-05-01T20:30:24.441Z",
"2016-03-31T20:30:24.441Z",
"2016-03-02T20:30:24.441Z",
"2016-01-31T20:30:24.441Z",
"2015-12-31T20:30:24.441Z",
"2015-12-01T20:30:24.441Z",
"2015-10-31T20:30:24.441Z",
"2015-10-01T20:30:24.441Z"
]
As you can see, this returns July twice, May twice, March twice, December twice, and October twice.
I actually just need the prior 12 months. The list I'd like is:
[
"2016-08-01T00:00:00.000Z",
"2016-07-01T00:00:00.000Z",
"2016-06-01T00:00:00.000Z",
"2016-05-01T00:00:00.000Z",
"2016-04-01T00:00:00.000Z",
"2016-03-01T00:00:00.000Z",
"2016-02-01T00:00:00.000Z",
"2016-01-01T00:00:00.000Z",
"2015-12-01T00:00:00.000Z",
"2015-11-01T00:00:00.000Z",
"2015-10-01T00:00:00.000Z",
"2015-09-01T00:00:00.000Z"
]
How in AQL could I ensure I always get the actual previous month instead of simply 30 days in the past? I fear that the date_subtract function won't handle leap years or 31sts.
It's hacky, but I managed to accomplish what I was after with this:
for mo in 0..11
return date_subtract(concat(left(date_iso8601(date_now()),7),'-01T00:00:00.000Z'), mo, "month")
Resulting in:
[
"2016-08-01T00:00:00.000Z",
"2016-07-01T00:00:00.000Z",
"2016-06-01T00:00:00.000Z",
"2016-05-01T00:00:00.000Z",
"2016-04-01T00:00:00.000Z",
"2016-03-01T00:00:00.000Z",
"2016-02-01T00:00:00.000Z",
"2016-01-01T00:00:00.000Z",
"2015-12-01T00:00:00.000Z",
"2015-11-01T00:00:00.000Z",
"2015-10-01T00:00:00.000Z",
"2015-09-01T00:00:00.000Z"
]
I would love a simpler solution to this, so please let me know if there is one. All those functions are a bunch of overhead.
Related
If a number is 2017 and I divide it by 4, I get 504.25.
if the number is 2016 an I divide it by 4, I get 504.
My question is, is there a way to use perl to check if a number that was divided, if the answer is a whole number and has NO decimal at all?
UPDATE: I got it working from the answer I selected below. I was trying to check if this is leap year, because today my code broke, because my code ran from the cron job as set to run on the last day of every month. Typically for feb, that is the 28th, so it ran on the 28th and shut the website done for the end of month maintenance. so what I have it do now is run on both Feb 28th and Feb 29th (For if there is a 29th) and in the code, I have it check if it is leap year now like this:
my $_leapNum = 4; # Leap year is always divisible by 4 with no remainder...
my $_NotLeapRemain = $_lTimeYYYY % $_leapNum; # I have $_lTimeYYYY built for the year only - longyear
if($_NotLeapRemain) {
my $_isLeapYear = 0;
} else {
my $_isLeapYear = 1;
}
Then later I check if it is leap year. If it is, then I check if the day is 28th, if it is, then it just exits. If it is the 29th, then it does what it was intended to do.
I tested it and it works perfectly now.
Thank you for your help.
-Rich
If you want to know if the result of dividing a $dividend by a $divisor is going to be a whole number or a number with a fractional portion, you can test that condition first:
if (my $remainder = $dividend % $divisor) {
print "$dividend cannot be divided evenly by $divisor.",
" There is a remainder of $remainder.\n";
}
else {
print "$dividend is evenly divisible by $divisor,",
" and the result is ", $divident / $divisor, "\n";
}
*EDIT: * This answer was posted in response to the original, un-edited question, which asked how to determine if the result of division contains a decimal portion. Now that the sense of the question changed toward date math, I agree that a DateTime solution is much better.
Using os.time how can I get how many months have passed since the unix epoch (Unix Timestamp)
I just need it for a month ID, so any kind of number would be fine, as long as it changes every month, and it can be reversed to get the actual month.
local function GetMonth(seconds)
local dayduration,year = 3600*24
local days={31,0,31,30,31,30,31,31,30,31,30,31}
for i=1970,10000 do -- For some reason too lazy to use while
local yeardays = i%4 == 0 and i%100 ~= 0 and 366 or 365
local yearduration = dayduration * yeardays
if yearduration < seconds then
seconds = seconds - yearduration
else
year = i break
end
end
days[2]=(year%4==0) and 29 or 28
seconds = seconds%(365*24*3600)
for i=1,12 do
if seconds>days[i]*dayduration then
seconds=seconds-days[i]*dayduration
else
return --i + year*12 <-- If you want a unique ID
end
end
end
Currently, it'll give the number 2, since it's February. If you uncomment the code at the end for the unique ID, you'll get 554 instead, meaning we're currently at the 554th month since the epoch.
As Jean-Baptiste Yunès said in his answer's comments, I'm not sure if your sentence:
NOTE: This is for Lua, but I'm unable to use os.date
meant you have no os.date, or that you don't know how to use it. You have an answer for both cases, you can use the one you need.
This may do the trick:
print (os.date("*t",os.time())["month"])
os.time() gives you the current date as a number. os.date("*t",...) converts it into a table in which the month equals to the number of the month corresponding to the date.
NOTE: I am a new Swift programmer, a NOOB if you will.
I am creating a school timetable app just for personal use to practise my coding. However, our school operates on a two week time table system, with 10 days, labeled 1 through to ten. I am wondering if anyone had some ideas as to how I could work out whether the current date is day one or day nine or day 4. I know I could use if statements for the dates, but the would take a long time, and require manual input of the dates. How could I have the app keep count of what day it is, skipping weekends?
EDIT - I could maybe have 14 days, with days 6,7,13 and 14 empty.
FOR EXAMPLE:
The current date is OCT 4, this is day one. I would like the app to be able to work out what day of the timetable the current date is. This would then load the appropriate day (e.g. Subject, Teacher, Classroom). Day One is Monday, Two is Tuesday, Five is Friday, Six is Monday, 10 is Friday. Could I have some sort of rostering system?
I am sorry if the question is vague, please tell me if I need to clarify.
I have been working on a fix for weeks now, so I have decided to turn to help. Any guidance whatsoever would be much appreciated, as I am at a dead end!
Many thanks
The numbers that I'm plugging into this example probably don't match your requirements but consider this as a strategy. (In this case, using a 1-to-14 cycle. If you'd rather get 1-to-10 you can put in a subtraction and a different error to throw on the "bad" days.)
class CyclicDay {
enum CyclicDayError: ErrorType {
case InvalidStartDate
}
lazy var baseline: NSDate? = {
// Set up some start date that matches day 1
var components = NSDateComponents()
components.day = 6
components.month = 9
components.year = 2015
return NSCalendar.currentCalendar().dateFromComponents(components)
}()
func dayOfCycle(testDate: NSDate) throws -> Int {
if let start = baseline {
// Convert difference to days
let interval = testDate.timeIntervalSinceDate(start)
let days = interval / (60 * 60 * 24)
// Convert to value 1..14 to position in a 2-week cycle
return Int(days % 14) + 1
}
throw CyclicDayError.InvalidStartDate
}
}
// Test today
let cd = CyclicDay()
let day = try cd.dayOfCycle(NSDate())
I would like to check if a given date is the first, second, third, fourth or last monday, tuesday, wednesday,..., sunday in the dates month.
The function should look like this:
if(checkdate(date, "second", "monday"))
{
alert("This is the second monday of the month");
}
else
{
alert("This is NOT the second monday of the month");
}
I would rather write a function that returns an object telling me the day and the week-number-in-month of the given date. Something like:
function infoDate(date) {
return {
day: date.getDay()+1,
week: (date.getDate()-1)%7+1
}
}
Then I can read the returned object and find out if it's the second monday or whatever:
var info = infoDate(date);
if(info.day==1 && info.week==2) {
// second monday
}
Of course, you can still write another localized function that does exactly what you ask for based on an array of numeral and day names.
use getDate() to get day of the month
then getDay() to get the day of the week
using these two you should be able to accomplish your task. Refer to this http://www.w3schools.com/jsref/jsref_obj_date.asp
It seems to me you only need to divide the date by 7 to know how many full weeks have passed, no matter what day it is. Subtracting one before dividing and adding it after sets the first week from 0 to one.
Date.prototype.weekofMonth= function(){
return Math.floor((this.getDate()-1)/7)+1;
}
alert(new Date().weekofMonth())
If I've got a time_t value from gettimeofday() or compatible in a Unix environment (e.g., Linux, BSD), is there a compact algorithm available that would be able to tell me the corresponding week number within the month?
Ideally the return value would work in similar to the way %W behaves in strftime() , except giving the week within the month rather than the week within the year.
I think Java has a W formatting token that does something more or less like what I'm asking.
[Everything below written after answers were posted by David Nehme, Branan, and Sparr.]
I realized that to return this result in a similar way to %W, we want to count the number of Mondays that have occurred in the month so far. If that number is zero, then 0 should be returned.
Thanks to David Nehme and Branan in particular for their solutions which started things on the right track. The bit of code returning [using Branan's variable names] ((ts->mday - 1) / 7) tells the number of complete weeks that have occurred before the current day.
However, if we're counting the number of Mondays that have occurred so far, then we want to count the number of integral weeks, including today, then consider if the fractional week left over also contains any Mondays.
To figure out whether the fractional week left after taking out the whole weeks contains a Monday, we need to consider ts->mday % 7 and compare it to the day of the week, ts->wday. This is easy to see if you write out the combinations, but if we insure the day is not Sunday (wday > 0), then anytime ts->wday <= (ts->mday % 7) we need to increment the count of Mondays by 1. This comes from considering the number of days since the start of the month, and whether, based on the current day of the week within the the first fractional week, the fractional week contains a Monday.
So I would rewrite Branan's return statement as follows:
return (ts->tm_mday / 7) + ((ts->tm_wday > 0) && (ts->tm_wday <= (ts->tm_mday % 7)));
If you define the first week to be days 1-7 of the month, the second week days 8-14, ... then the following code will work.
int week_of_month( const time_t *my_time)
{
struct tm *timeinfo;
timeinfo =localtime(my_time);
return 1 + (timeinfo->tm_mday-1) / 7;
}
Assuming your first week is week 1:
int getWeekOfMonth()
{
time_t my_time;
struct tm *ts;
my_time = time(NULL);
ts = localtime(&my_time);
return ((ts->tm_mday -1) / 7) + 1;
}
For 0-index, drop the +1 in the return statement.
Consider this pseudo-code, since I am writing it in mostly C syntax but pretending I can borrow functionality from other languages (string->int assignment, string->time conversion). Adapt or expand for your language of choice.
int week_num_in_month(time_t timestamp) {
int first_weekday_of_month, day_of_month;
day_of_month = strftime(timestamp,"%d");
first_weekday_of_month = strftime(timefstr(strftime(timestamp,"%d/%m/01")),"%w");
return (day_of_month + first_weekday_of_month - 1 ) / 7 + 1;
}
Obviously I am assuming that you want to handle weeks of the month the way the standard time functions handle weeks of the year, as opposed to just days 1-7, 8-13, etc.