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.
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.
I have to develop a system, where user will specify a starting range and an ending range (By Range, I mean to say a particular period, where PERIOD is given as a concatenation of YEAR AND MONTH). For example, PERIOD = 201304, where 2013 is the user entered Year and 04 is the MONTH. The user can specify a maximum range of upto 2 years only.
Data needs to be selected on the basis of the user entered range. The problem is whenever I try to loop through the period, the PERIOD changes after 201312 to 201313. I have separate variables for user selected year and month (start_year, start_month, end_year, end_month)
I did a IF loop there in which I tried to do the following
FOR p_tmpyear = p_tempfrom TO p_tempto
IF (p_monthfrm < 12) THEN
LET p_yearfrm = p_yearfrm + 1
LET p_monthfrm = 01
LET p_fromperiod = p_yearfrm + 1,p_monthfrm >p_fromperiod is an integer storing concatenated Month and Year, to achieve the desired PERIOD format as mentioned above.
LET p_tempfrom = p_fromperiod
END IF
DISPLAY p_tmpyear
END FOR
I even tried thsi one :
IF (p_fromperiod MOD p_yearfrm = 13) THEN
LET p_yearfrm = p_yearfrm + 1
LET p_monthfrm = 01
LET p_fromperiod = p_yearfrm + 1,p_monthfrm
Still the period changes after reaching 201212 to 201213. I want this to be 201301. Please help.
As you have probably guessed, you have some flaws in your logic. The first example is too hard to decipher, whilst the second shows you don't understand what the MOD operator does. If used correctly, in this context, it would be of the form IF variable MOD 12 = 0 i.e. every 12th value do something.
I think where you are going wrong is trying to do it all in one loop. I would keep it simple and break the problem into two nested loops, one for year, one for month
DEFINE start_year, end_year, start_month, end_month INTEGER -- sample values 2012, 2013,7,6
DEFINE loop_year, loop_month INTEGER -- values used to loop through year and month respectively
DEFINE loop_first_month, loop_last_month INTEGER -- for each year, calc first month and last month
DEFINE period CHAR(6)
-- First loop over year
FOR loop_year = start_year TO end_year
-- Calculate first and last month for the given year
-- If first year, use passed in start month else use 1
IF loop_year = start_year THEN
LET loop_first_month = start_month
ELSE
LET loop_first_month = 1
END IF
-- If last year, use passed in end month else use 12
IF loop_year = end_year THEN
LET loop_last_month = end_month
ELSE
LET loop_last_month = 12
END IF
-- Second loop over month
FOR loop_month = loop_first_month TO loop_last_month
LET period = loop_year USING "&&&&",loop_month USING "&&"
DISPLAY period
END FOR
END FOR
I would also encourage you to also post questions like this in the dedicated 4Js Genero developers forum at the 4Js website http://www.4js.com/fjs_forum/ as well as here. Most Genero developers should be aware of that forum and have access.
I'm attempting to pull some information out of text file that is updated after I query a piece of equipment. The text file contains lines such as shown here (abbreviated):
05-Nov-13 11:11:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:56.3496 ( -1 7020 10244) scpeng.exe:CServer.cpp,....
The text file can contain up to several weeks of information. I have a subroutine that will run a few seconds after I query the equipment which should allow for the reply and the applicable line to be present in the text file. In the routine, I am trying to scroll through the lines examining the date to arrive at the date of the subroutine call followed by the time (or a time ~10 seconds prior the the current time) to arrive at the lines pertinent to where the information could be found.
do
msg = msgstream.ReadLine
logdate = mide(msg,1,9)
logday = Cdate(logdate)
loop while logday < date
do
msg = msgstream.Readline
logtime = mid(msg,12,8)
'logtime = CDate(logtime) This mod is not working
loop while logtime < time
The date loop appears to work however the time is giving me problems. It does not error out but I can't get it to run beyond one line of text. Can anyone suggest a fix or better option? I have read that the built-in Date function can include the time but I do not believe this version I'm using does. Also, the text file contains times in a 24 hour format where I believe the time function returns values in a 12 hr format ie "12:43:27 PM ST".
You're making this way too complicated. Simply parse the whole date string into a datetime value:
refdate = Now
Do
msg = msgstream.ReadLine
logdate = CDate(Mid(msg, 1, 19))
Loop While logdate < refdate
You can extract date and time portions from the value later, e.g. like this:
WScript.Echo DateValue(logdate)
WScript.Echo TimeValue(logdate)
Also, Time returns the current (unformatted) system time. Whether it's displayed in 12 hour or 24 hour format depends on your system's region settings. However, you can always get the hour (0-23) by using the Hour function.
Parse each line with a regex to get the correct date and time part. I prefer a regexp above string manipulation functions because you can separate format and code.
Reassemble the date from the two parts and see if the date is smaller than yesterday at this time.
Option Explicit
dim strTest, re, matches, myDatePart, myTimePart, logDate
' teststring
strTest = "08-Nov-13 14:10:56.3496 ( -1 7020 10244) scpeng.exe:CServer.cpp,...."
Set re = new regexp
' This pattern extracts two part, the date as (dd-www-dd) and the time as (hh:mm:ss)
re.pattern = "(\d{2}-\w{3}-\d{2}) +(\d{2}:\d{2}:\d{2})"
Set matches = re.Execute(strTest)
' Get the first and second submatch to define the date and time
myDatePart = matches(0).submatches(0)
myTimePart = matches(0).submatches(1)
' datevalue and timevalue automatically tranforms to Date type
logDate = datevalue(myDatePart) + timevalue(myTimePart)
' See if the date is smaller than yesterday exactly this time
msgbox (logDate < (DateAdd("d", -1, now))) ' Returns True, because 08 Nov is earlier than yesterday.
I simply want to generate a series of dates 1 year apart from today.
I tried this
CurveLength=30;
t=zeros(CurveLength);
t(1)=datestr(today);
x=2:CurveLength-1;
t=addtodate(t(1),x,'year');
I am getting two errors so far?
??? In an assignment A(I) = B, the number of elements in B and
Which I am guessing is related to the fact that the date is a string, but when I modified the string to be the same length as the date dd-mmm-yyyy i.e. 11 letters I still get the same error.
Lsstly I get the error
??? Error using ==> addtodate at 45
Quantity must be a numeric scalar.
Which seems to suggest that the function can't be vectorised? If this is true is there anyway to tell in advance which functions can be vectorised and which can not?
To add n years to a date x, you do this:
y = addtodate(x, n, 'year');
However, addtodate requires the following:
x must be a scalar number, not a string.
n must be a scalar number, not a vector.
Hence the errors you get.
I suggest you use a loop to do this:
CurveLength = 30;
t = zeros(CurveLength, 1);
t(1) = today; % # Whatever today equals to...
for ii = 2:CurveLength
t(ii) = addtodate(t(1), ii - 1, 'year');
end
Now that you have all your date values, you can convert it to strings with:
datestr(t);
And here's a neat one-liner using arrayfun;
datestr(arrayfun(#(n)addtodate(today, n, 'year'), 0:CurveLength))
If you're sequence has a constant known start, you can use datenum in the following way:
t = datenum( startYear:endYear, 1, 1)
This works fine also with months, days, hours etc. as long as the sequence doesn't run into negative numbers (like 1:-1:-10). Then months and days behave in a non-standard way.
Here a solution without a loop (possibly faster):
CurveLength=30;
t=datevec(repmat(now(),CurveLength,1));
x=[0:CurveLength-1]';
t(:,1)=t(:,1)+x;
t=datestr(t)
datevec splits the date into six columns [year, month, day, hour, min, sec]. So if you want to change e.g. the year you can just add or subtract from it.
If you want to change the month just add to t(:,2). You can even add numbers > 12 to the month and it will increase the year and month correctly if you transfer it back to a datenum or datestr.
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.