Is there a way to find number of days from timespan?
For example,
time(00:00:00.2000000), time(00:30:30), time(01:00:00), time(413.00:00:00)
should return 0, 0, 0, 413
You could use format_timespan():
let getDays = (t:timespan)
{
toint(format_timespan(t, 'd'))
};
print result = getDays(time(00:00:00.2000000)), //0
getDays(time(00:30:30)), //0
getDays(time(01:00:00)), //0
getDays(time(413.00:00:00)) //413
An alternative way would be to divide the timespan by the a day, for example:
datatable(t:timespan) [ time(00:00:00.2000000), time(00:30:30), time(01:00:00), time(413.00:00:00)]
| extend Days = tolong(t/1d)
Related
I have a timestamp (submitTime) which I need to check whether it is less than 1 hour old or not. Timestamps are in microseconds and including date.
currentTime = 1527530605357000000 (Monday, May 28, 2018 6:03:25.357 PM)
submitTime = 1527529918658907821 (Monday, May 28, 2018 5:51:58.659 PM)
long currentTime = (long) (new Date().getTime()*1000000)
submitTime = job.SubmitTime // part of the code
oneHhour = 3600000000
if (currentTime - submitTime > oneHhour) {
println job.Name + " env is up more than 1 hour";
But it doesn't work since the result is 686698092179 and it it not represent time.
Help?
Assuming SubmitTime is a timestamp in microseconds, you can compare it the the current timestamp in microseconds like so:
// Get the current time (System.currentTimeMillis) in microseconds:
long currentMicroseconds = TimeUnit.MILLISECONDS.toMicros(System.currentTimeMillis())
// You could also simply do this:
long currentMicroseconds = System.currentTimeMillis() * 1000
// Subtract the timestamps and compare:
if (currentMicroseconds - job.SubmitTime > 3600000000) {
// More than an hour has elapsed
}
The timestamp is assumed to be the number of microseconds since January 1, 1970, 00:00:00 GMT (consistent with Date.getTime).
In groovy you can use TimeCategory which is much more intuitive:
def date = new Date(timestampInLong)
use (groovy.time.TimeCategory) {
println (date > new Date() - 1.hour)
}
We have a solution in Delphi that calculates a travel's duration of a given vehicle, for example, 20 minutes, 25 minutes and so on. However, sometimes we have to antecipate the travel's start time, from a specific datetime, for example 09:00 to 08:40. Then, we need to substract a negative value from a TDateTime variable (travel's start), in this case, something like "-00:20". To do this, we multiply the datetime value by -1 (for example MyDiffDateTimeVariable * -1). The output we got is very strange, sometimes we obtain the exactly opposite behavior. In other case, an operation to extract 20 minutes results in a difference of two days from the original datetime.
Here is a sample console application that simulate our situation, with the current outputs, and what we will expected:
program DateTimeSample;
uses
System.SysUtils, System.DateUtils;
var
LDate1: TDateTime;
LDate2: TDateTime;
begin
LDate1 := IncMinute(0, 20);
LDate2 := IncMinute(0, -20);
WriteLn('Date1: ' + DateTimeToStr(LDate1));
// Output = Date1: 30/12/1899 00:20:00 [OK]
WriteLn('Date2: ' + DateTimeToStr(LDate2));
// Output = Date2: 29/12/1899 23:40:00 [OK]
WriteLn('-----');
WriteLn('Date1: ' + DateTimeToStr(LDate1 * -1));
// Output = Date1: 30/12/1899 00:20:00 [Expected 29/12/1899 23:40:00]
WriteLn('Date2: ' + DateTimeToStr(LDate2 * -1));
// Output = Date2: 31/12/1899 23:40:00 [Expected 30/12/1899 00:20:00]
ReadLn;
end.
When you inspect the value casted to double, you can see:
double(LDate1) = 0.0138888888888889
double(LDate2) = -1.98611111111111
Seems like a bug to me, because with today it returns:
double(LDate1) = 43168,0138888889
double(LDate2) = 43167,9861111111
Edit: Hmm, according the documentation, it is not a bug, it is a feature :-)
When working with negative TDateTime values, computations must handle time portion separately. The fractional part reflects the fraction of a 24-hour day without regard to the sign of the TDateTime value. For example, 6:00 A.M. on December 29, 1899 is –1.25, not –1 + 0.25, which would equal –0.75. There are no TDateTime values from –1 through 0.
Karel's answer explains what's happening. Basically, TDateTime is represented as a Double, but that doesn't mean you can work with it in the same way as you normally would a Double value. It's internal structure carries particular semantics that if you don't handle them correctly, you're bound to get some peculiar behaviour.
The key mistake you're making is in taking the negative of a date-time value. This concept doesn't really make sense. Not even if you look at dates in BC, because the calendar system has changed a number of times over the years.
This is the main reason you should favour library routines that deal with the nuances of the internal structure (whatever your platform). In Delphi that means you should use the SysUtils and DateUtils routines for working with dates and times.
You seem to be trying to hold duration as a TDateTime value. You'd be much better off determining your preferred unit of measure and using Integer (perhaps Int64) or Double (if you need support for fractions of a unit). Then you can add or subtract, preferably using library routines, the duration from your start or end times.
The following code demonstrates some examples.
var
LStartTime, LEndTime: TDateTime;
LDuration_Mins: Integer;
begin
{ Init sample values for each calculation }
LStartTime := EncodeDateTime(2018, 3, 9, 8, 40, 0, 0);
LEndTime := EncodeDateTime(2018, 3, 9, 9, 0, 0, 0);
LDuration_Mins := 20;
{ Output result of each calculation }
Writeln(Format('Whole Duration: %d', [MinutesBetween(LStartTime, LEndTime)]));
Writeln(Format('Frac Duration: %.6f', [MinuteSpan(LStartTime, LEndTime)]));
Writeln(Format('Start Time: %s', [FormatDateTime('yyyy-mm-dd hh:nn:ss', IncMinute(LEndTime, -LDuration_Mins))]));
Writeln(Format('End Time: %s', [FormatDateTime('yyyy-mm-dd hh:nn:ss', IncMinute(LStartTime, LDuration_Mins))]));
end;
Additional Considerations
You said you're dealing with vehicle travel times. If you're dealing with long-haul travel you might have some other things to think about.
Daylight saving: If a vehicle starts its journey shortly before DST changes and ends after, you need to take this into account when calculating a missing value. Perhaps easiest would be to convert date-time values to UTC for the calculation. Which leads to...
Time zone changes: Again, unless your code is time-zone aware you're bound to make mistakes.
Compiler always appears to treat TDateTime as positive when doing numerical operations on it. Try this:
uses
System.SysUtils, System.DateUtils;
function InvertDate(ADateTime: TDateTime): TDateTime;
var
LMsec: Int64;
begin
LMsec := MillisecondsBetween(ADateTime, 0); //Always Positive
if ADateTime > 0 then
LMsec := 0 - LMsec;
Result := IncMillisecond(0, LMsec);
end;
var
LDate1: TDateTime;
LDate1Negative: TDateTime;
LDate2: TDateTime;
begin
try
LDate1 := IncMinute(0, 20);
LDate2 := IncMinute(0, -20);
WriteLn('Date1: ' + DateTimeToStr(LDate1));
// Output = Date1: 30/12/1899 00:20:00 [OK]
WriteLn('Date2: ' + DateTimeToStr(LDate2));
// Output = Date2: 29/12/1899 23:40:00 [OK]
WriteLn('-----');
WriteLn('Date1: ' + DateTimeToStr( InvertDate(LDate1) ));
// Output = Date1: Expected 29/12/1899 23:40:00
WriteLn('Date2: ' + DateTimeToStr( InvertDate(LDate2) ));
// Output = Date2: 30/12/1899 00:20:00
ReadLn;
except
on E: Exception do
Writeln(E.ClassName, ': ', E.Message);
end;
end.
I am very new to lua and my plan is to create a table. This table (I call it test) has 200 entries - each entry has the same subentries (In this example the subentries money and age):
This is a sort of pseudocode:
table test = {
Entry 1: money=5 age=32
Entry 2: money=-5 age=14
...
Entry 200: money=999 age=72
}
How can I write this in lua ? Is there a possibility ? The other way would be, that I write each subentry as a single table:
table money = { }
table age = { }
But for me, this isn't a nice way, so maybe you can help me.
Edit:
This question Table inside a table is related, but I cannot write this 200x.
Try this syntax:
test = {
{ money = 5, age = 32 },
{ money = -5, age = 14 },
...
{ money = 999, age = 72 }
}
Examples of use:
-- money of the second entry:
print(test[2].money) -- prints "-5"
-- age of the last entry:
print(test[200].age) -- prints "72"
You can also turn the problem on it's side, and have 2 sequences in test: money and age where each entry has the same index in both arrays.
test = {
money ={1000,100,0,50},
age={40,30,20,25}
}
This will have better performance since you only have the overhead of 3 tables instead of n+1 tables, where n is the number of entries.
Anyway you have to enter your data one way or another. What you'd typically do is make use some easily parsed format like CSV, XML, ... and convert that to a table. Like this:
s=[[
1000 40
100 30
0 20
50 25]]
test ={ money={},age={}}
n=1
for balance,age in s:gmatch('([%d.]+)%s+([%d.]+)') do
test.money[n],test.age[n]=balance,age
n=n+1
end
You mean you do not want to write "money" and "age" 200x?
There are several solutions but you could write something like:
local test0 = {
5, 32,
-5, 14,
...
}
local test = {}
for i=1,#test0/2 do
test[i] = {money = test0[2*i-1], age = test0[2*i]}
end
Otherwise you could always use metatables and create a class that behaves exactly like you want.
I have a date/time field from a shopping cart API feed, but I don't know what format it is in and I don't have access to the database.
What could [1252457867] be for a date?
These dates are all within the last couple weeks
Any ideas?
Clearly a unix timestamp.
1252457867 = 09 Sep 2009 - 02:57:47
This sounds like seconds since the Unix Epoch (January 1, 1970).
That looks like seconds elapsed since Jan. 1st, 1970 12:00AM.
Use this function to get the date:
var baseDate = new DateTime(1970, 1, 1, 0, 0, 0);
var transactionDate = baseDate.AddSeconds(1252457867);
This will output {9/9/2009 12:57:47 AM} PST
**EDIT: **
If you need UTC:
var utcDate = baseDate.AddSeconds(1252457867).ToUniversalTime();
This outputs {9/9/2009 7:57:47 AM}
--Adam
Flex is driving me CRAZY and I think it's some weird gotcha with how it handles leap years and none leap years. So here's my example. I have the below dateDiff method that finds the number of days or milliseconds between two dates. If I run the following three statements I get some weird issues.
dateDiff("date", new Date(2010, 0,1), new Date(2010, 0, 31));
dateDiff("date", new Date(2010, 1,1), new Date(2010, 1, 28));
dateDiff("date", new Date(2010, 2,1), new Date(2010, 2, 31));
dateDiff("date", new Date(2010, 3,1), new Date(2010, 3, 30));
If you were to look at the date comparisons above you would expect to get 30, 27, 30, 29 as the number of days between the dates. There weird part is that I get 29 when comparing March 1 to March 31. Why is that? Is it something to do with February only having 28 days? If anyone has ANY input on this that would be greatly appreciated.
public static function dateDiff( datePart:String, startDate:Date, endDate:Date ):Number
{
var _returnValue:Number = 0;
switch (datePart) {
case "milliseconds":
_returnValue = endDate.time - startDate.time;
break;
case "date":
// TODO: Need to figure out DST problem i.e. 23 hours at DST start, 25 at end.
// Math.floor causes rounding down error with DST start at dayOfYear
_returnValue = Math.floor(dateDiff("milliseconds", startDate, endDate)/(1000 * 60 * 60 * 24));
break;
}
return _returnValue;
}
This is not a leap year problem, but rather a daylight savings time problem.
To correct the code to account for DST, you need to look at the timezoneOffset of both dates to determine if the date range is spanning a DST boundary.
var adjustment:Number = ( startDate.timezoneOffset - endDate.timezoneOffset ) * 60 * 1000;
_returnValue = endDate.time - startDate.time + adjustment;
This will get the difference between the two time zones (in minutes), convert that value to milliseconds, and then apply the timezone difference to the millisecond difference to "cancel out" the DST boundary.
Naturally, when both numbers are in the same time zone, the adjustment value becomes 0 and the time values are not adjusted.
You have part of the answer in your comment: 2010-Mar-01 0:00 until 2010-Mar-31 0:00 is thirty (!) days minus one hour (because Mar 14 is DST start in 2010). Since you floor the result of your division, you get 29.
Edit: This answer is of course based on the assumption that the time property of Date takes DST into account. This would explain your problem; I didn't check it, however.