Maybe i am missing something very stupid so forgive me .
Comparing a date that is 1 month later than another date , starts to give me strange numbers.
I guess its because the result is float ? Here is how i do the comparison :
int daysToCheckUses=60;
long seconds=60*60*24*daysToCheckUses;
NSDate *today=[NSDate date];
if([today timeIntervalSinceDate:date]>seconds) //date can be more than 60 days old
Is there something wrong with this,when using big numbers? for example when the interval is 1 month i get 518400,but for 3 months i get 18662400000 i know that the comparisons returns float number that can't hold these numbers. I am afraid also to get a crash when a few months will pass.
This method to calculate a big distance is not that good, i found a great and better way for this using a great answer from here :
Number of days between two NSDates
Where the comparison is for days, than months will only give a few 100's integer .
Related
When I use the formula below the results of the EOMONTH function
returns the start of the next month for any month with 30 days instead of the last day of the specified month. The month and years are correct, so I'm pretty sure it's EOMONTH when used in another function.
For example,the results in B3 should be "11/31/1965" but it returns "12/1/1965".
=DATE(YEAR(B2),MONTH(B2)+6,DAY(TEXT(EOMONTH(MONTH(B2)+6,0))))
I have tried subtracting a day, but it returns the end-of-month -1 for months with 31 days (30). So I have the same problem in the other case.
I have also used IFS() to account for months with 30 days, and it miscalculates the date the same way.
=IFS( MONTH(B2)+6 = 4,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
6,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
9,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
MONTH(B2)+6 =
11,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0))-2) ,
TRUE ,DATE(YEAR(B2),MONTH(B2)+6,DAY(EOMONTH(MONTH(B2)+6,0)) ) )
The EOMONTH function by itself where I just pass in the date as a string works correctly (column F).
Any Idea on what I'm doing wrong?
Thanks in advance.
Yes, as #player0 has explained, you can't just add something to a month and feed it into eomonth. Try putting
=eomonth(month(B2)+6,0)
into B3 (formatted as a date).
You get
1/31/1900
Why? month(b2)+6 gives 11 (which is just a number). Dates in google sheets are represented as days since 12/31/1899. So 11 formatted as a date gives 1/11/1900. Applying eomonth to that gives the last day of January 1900, which is the 31st. Feeding that into your formula would give 11/31/65, but that date doesn't exist, so you get 12/1/65.
If you want to go forward 6 months and then get the last day of the month, you need
=eomonth(date(year(B2),month(B2)+6,1),0)
You can also use the Edate function, which does not roll over into the first day of the next month:
=eomonth(edate(B2,6),0)
EOMONTH does not understand MONTH. instead, it converts it into date. to use EOMONTH you need to supply it with valid date
=EOMONTH(B2, 0)
There's the subtract() method but the documentation says it's not aware of daylight savings which makes it pretty much useless in this case, or in any other case except where the programmer doesn't know how many milliseconds there are in 24 hours.
I'm thinking of two ways:
get the day of the month, and then subtract N from it and if it's less than 1 then subtract the month and the year if appropriate and set the day for the last day of whichever month it turns out to be
OR
subtract N days from the noon of the current day and then get start of the day for the resulting day
Is there some easier/better way to do this?
You should probably try to convert both DateTimes to UTC (standardize), then call difference(). That converts it to a nice, easy Duration, which you can convert as necessary to hours, days, months, or whatever else.
DateTime one = somedatetime.toUtc();
DateTime two = someotherdatetime.toUtc();
Duration diff = one.difference(two);
//Then just convert...
return diff.inDays;
I calculated a duration between two times, e.g. between 9:00 am and 11:00 am. So far so good. Now I need to decide if this duration is less 6 hours.
I do remember that this was pain in the s in excel but nevertheless I tried it the simple way:
=IF(E2 < 06:00:00; "y"; "n")
of course that didn't work. Next:
=IF(DURATION(E2) < DURATION(06:00:00); "y"; "n")
still, it didn't work.
So, okay, how can I compare two duration?
Divide hours by 24:
=IF(E2 < 6/24, "y", "n")
Value is E2 is a formatted time, actually 1 hour is 1/24, 1 day is 1.
Some info about date and time formats here:
http://www.excel-easy.com/examples/date-time-formats.html
You can also use the HOUR function if you want to
=if(HOUR(E2)<6,ʺyesʺ,ʺnoʺ)
or
=if(E2<time(6,0,0),ʺyesʺ,ʺnoʺ)
(if you write 06:00:00 in a formula it takes it as a string not a time)
but as I'm sure someone is about to point out, the first formula above gives the wrong answer for durations of more than a day (because it takes the hour part of a datetime).
What I find interesting is that you can assume for a worksheet formula that dates and times are represented as whole numbers (days) and fractions (parts of a day) just like in Excel. If you ever have to deal with them in Google App Scripts, you suddenly find that it's object-oriented and you have no choice but to use methods like hour() to manipulate them.
I needed to use the equivalent of:
=if(TIMEVALUE(E2)<6/24, "yes", "no")
I am getting lost and nuts with the DATETIME in Informix.I have two problems which I can hardly solve:
I have a DATETIME column (e.g. starttime) which I need to convert into an int value, e.g. seconds of year or epoch, or whatever. I found some conversion into utc, but this depends on the timezone the server runs, which I have no idea how to specify for the conversion.... Any other hint, how to convert it into seconds would be appreciated.
I need to calculate the difference between two DATETIME-fields (e.g. endtime-starttime) and then sum it up. To my understanding the result is interval day(13) to fraction(3). I need to convert the sum once again into seconds, cause I need to update other values with this result.
So, can anybody help me as to how to convert within a SQL-statement the different result-types?
CAST it to INT, example:
select
((current + 5 units day - current)::interval second(9) to second)
,((current + 5 units day - current)::interval second(9) to second)::char(10)::int8
from systables
where tabid=1
I have a start/end times for a calculation I'm trying to do and am having a problem seeing if the end time is before 12AM the day after the start time. Also, I need to calculate how many days past the start time it is.
What I have: Start Date, End Date
What I need:
- How many 'Midnights' is the End Date past the Start Date?
Has anyone done anything like this?
This uses PHP 5.3, if you have an earlier version you may need to use unix timestamps to figure out the difference. The number of midnights should be the number of days difference assuming both start and end times have the same time. So setting both to be midnight of their current day setTime(0,0), should make the calculation correct.
Using the DateTime objects.
$start = new DateTime('2011-03-07 12:23:45');
$end = new DateTime('2011-03-08 1:23:45');
$start->setTime(0,0);
$end->setTime(0,0);
$midnights = $start->diff($end)->days;
Without using the setTime() calls, this would result in 0, because there is less than 24 hours between start and end. With the setTime() this results in 1 because now the difference is exactly 24 hours.
The diff() function was introduced in 5.3 along with the DateInterval class. In 5.2 you can still use the DateTime class but will have to work out the total days using the Unix timestamp.
$midnights = ($end->format('U') - $start->format('U')) / 86400
You can wrap that in an abs() function to the order of start/end does not matter.
Note: These functions may need to be tested for cases that involve DST.
A comment in the php date documentation uses round after dividing by 86400 (number of seconds in a day), to counter any issues that could be involved with DST.
An alternative approach with DateTimes would be to create them in the UTC.
$utcTimezone = new DateTimeZone('UTC');
$start = new DateTime('2011-03-07 12:23:45', $utcTimezone);
$end = new DateTime('2011-03-08 1:23:45', $utcTimezone);