I need to be able to compare the number of whole days between two dates in ActionScript, is this possible?
I'd like to test if one date is 7 days or less after today, and if so is it one day or less (if it's before today this also counts).
The workaround I have in place is using the .time part of the date field:
// Get the diffence between the current date and the due date
var dateDiff:Date = new Date();
dateDiff.setTime (dueDate.time - currentDate.time);
if (dateDiff.time < ( 1 * 24 * 60 * 60 * 1000 ))
return "Date is within 1 day");
else if (dateDiff.time < ( 7 * 24 * 60 * 60 * 1000 ))
return "Date is within 7 days");
As I say - this is only a workaround, I'd like a permanent solution to allow me to check the number of whole days between 2 dates. Is this possible?
Thanks
var daysDifference:Number = Math.floor((dueDate.time-currentDate.time)/(1000*60*60*24));
if (daysDifference < 2)
return "Date is within 1 day";
else if (daysDifference < 8)
return "Date is within 7 days";
Related
I would like to have a Firestore security rule for bithdate field in my users collections where user age should be >=18 and <=80. I tried the following rule but I know it is not going to work especially for ages that are close to 18 or 80. Any idea how to make this work.
let now = request.time;
let thisYear = now.year();
let thisMonth = now.month();
let thisDay = now.day();
request.resource.data.birthdate >= timestamp.date(thisYear-80,thisMonth,thisDay) &&
request.resource.data.birthdate <= timestamp.date(thisYear-18,thisMonth,thisDay)
You can achieve the same result by comparing it with the request time but in seconds. here’s one example on how this will work:
function isAbove18(date) {
return date.seconds <= request.time.seconds - 18 * 365.25 * 24 * 60 * 60; //considering leap year with 0.25 🙃
}
function isBelow80(date) {
return date.seconds >= request.time.seconds - 80 * 365.25 * 24 * 60 * 60;
}
allow read, write: if isAbove18(resource.data.birthdate) && isBelow80(resource.data.birthdate);
For more about this you can go through this docs which takes some different approaches.
If I have a date YYMMDDHHmmss such as 190525234530 how do I work out the smallest number of characters to represent this using 0-9a-z (36 characters)?
I believe there are 3,153,600,000 combinations (100 years * 365 days * 24 hours * 60 minutes * 60 seconds) which would fit into 32 bits. Does this mean I could represent these dates using 4 characters?
I am a bit lost as to how to do the conversion so if anyone could show me the maths that would be greatly appreciated.
I ended up doing this in JavaScript, I decided I wanted to compress to 6 characters so I created my own time which generates unique ID's for up to 68 years from 01/01/2019 which worked for me.
function getId() {
//var newTime = parseInt(moment().format("X")) - 1546300800;//seconds since 01/01/2019
var newTime = Math.floor((new Date()).getTime() / 1000) - 1546300800;//seconds since 01/01/2019
var char = "0123456789abcdefghijklmnopqrstuvwxyz";//base 36
return char[Math.floor(newTime / 36**5)]+
char[Math.floor(newTime%36**5 / 36**4)]+
char[Math.floor(newTime%36**4 / 36**3)]+
char[Math.floor(newTime%36**3 / 36**2)]+
char[Math.floor(newTime%36**2 / 36)]+
char[Math.floor(newTime%36)];
}
console.log(getId());
Thanks to #user956584 this can be be changed to:
function getId() {
//var newTime = parseInt(moment().format("X")) - 1546300800;//seconds since 01/01/2019
var newTime = Math.floor((new Date()).getTime() / 1000) - 1546300800;//seconds since 01/01/2019
return newTime.toString(36);
}
console.log(getId());
I would like to find out the number of hours and minutes between two date time stamp.
if for example
sDateTime = 2016-01-01 01:00
eDateTime = 2016-01-03 02:30
I would like it to output it as 49:30 (49hours and 30minutes)
I am unable to figure a method to work this out.
what I have so far:
Set oMNOF=##class(MNOF.MNOF).%OpenId(Id)
Set zstartDt=oMNOF.sDateTime
Set startDt=$PIECE(zstartDt,",",1)
Set startTime=$PIECE(zstartDt,",",2)
Set zendDt=oMNOF.eDateTime
Set endDt=$PIECE(zendDt,",",1)
Set endTime=$PIECE(zendDt,",",2)
set dateDiff=((endDt - startDt)) //2 days
set timeDiff=(endTime - startTime) //outputs 5400 seconds
set d = (dateDiff * 24 * 60 * 60)
set h = ((timeDiff - d) / 60)
set m = timeDiff - (d) - (h * 60)
Thank you for the help.
Another option:
USER>set mm=$system.SQL.DATEDIFF("mi","2016-01-02 01:00","2016-01-03 02:30")
USER>write "hours=", mm \ 60
hours=25
USER>write "minutes=", mm # 60
minutes=30
Hi thanks to all for the help.
I managed to come up with the below, appreciate if someone can improve on this.
<script language="cache" method="MGetData" arguments="pStartDt:%String,pEndDt:%String,pTimeField:%String" returntype="%Library.String">
set val1="00"
//HOUR: check if length equals 1
if $LENGTH($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600))=1{
//add leading zero
set val1 ="0"_$SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600)
}
else{
//get without leading zero
set val1 = $SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600)
}
//MINUTES: check if length equals 1
if $LENGTH($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/60) - ($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600)*60))=1{
//add leading zero
set val2 ="0"_($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/60) - ($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600)*60))
}
else{
//get without leading zero
set val2 = ($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/60) - ($SYSTEM.SQL.FLOOR($system.SQL.DATEDIFF("ss",pStartDt,pEndDt)/3600)*60))
}
//insert result data into the time field
Write "document.getElementById('"_pTimeField_"').value='"_val1_":"_val2_"';"
//Write "alert('"_val1_"^"_val2_"');"
QUIT 1
I have a variable with a date table that looks like this
* table:
[day]
* number: 15
[year]
* number: 2015
[month]
* number: 2
How do I get the days between the current date and the date above? Many thanks!
You can use os.time() to convert your table to seconds and get the current time and then use os.difftime() to compute the difference. see Lua Wiki for more details.
reference = os.time{day=15, year=2015, month=2}
daysfrom = os.difftime(os.time(), reference) / (24 * 60 * 60) -- seconds in a day
wholedays = math.floor(daysfrom)
print(wholedays) -- today it prints "1"
as #barnes53 pointed out could be off by one day for a few seconds so it's not ideal, but it may be good enough for your needs.
You can use the algorithms gathered here:
chrono-Compatible Low-Level Date Algorithms
The algorithms are shown using C++, but they can be easily implemented in Lua if you like, or you can implement them in C or C++ and then just provide Lua bindings.
The basic idea using these algorithms is to compute a day number for the two dates and then just subtract them to give you the number of days.
--[[
http://howardhinnant.github.io/date_algorithms.html
Returns number of days since civil 1970-01-01. Negative values indicate
days prior to 1970-01-01.
Preconditions: y-m-d represents a date in the civil (Gregorian) calendar
m is in [1, 12]
d is in [1, last_day_of_month(y, m)]
y is "approximately" in
[numeric_limits<Int>::min()/366, numeric_limits<Int>::max()/366]
Exact range of validity is:
[civil_from_days(numeric_limits<Int>::min()),
civil_from_days(numeric_limits<Int>::max()-719468)]
]]
function days_from_civil(y, m, d)
if m <= 2 then
y = y - 1
m = m + 9
else
m = m - 3
end
local era = math.floor(y/400)
local yoe = y - era * 400 -- [0, 399]
local doy = math.modf((153*m + 2)/5) + d-1 -- [0, 365]
local doe = yoe * 365 + math.modf(yoe/4) - math.modf(yoe/100) + doy -- [0, 146096]
return era * 146097 + doe - 719468
end
local reference_date = {year=2001, month = 1, day = 1}
local date = os.date("*t")
local reference_days = days_from_civil(reference_date.year, reference_date.month, reference_date.day)
local days = days_from_civil(date.year, date.month, date.day)
print(string.format("Today is %d days into the 21st century.",days-reference_days))
os.time (under Windows, at least) is limited to years from 1970 and up. If, for example, you need a general solution to also find ages in days for people born before 1970, this won't work. You can use a julian date conversion and subtract between the two numbers (today and your target date).
A sample julian date function that will work for practically any date AD is given below (Lua v5.3 because of // but you could adapt to earlier versions):
local
function div(n,d)
local a, b = 1, 1
if n < 0 then a = -1 end
if d < 0 then b = -1 end
return a * b * (math.abs(n) // math.abs(d))
end
--------------------------------------------------------------------------------
-- Convert a YYMMDD date to Julian since 1/1/1900 (negative answer possible)
--------------------------------------------------------------------------------
function julian(year, month, day)
local temp
if (year < 0) or (month < 1) or (month > 12)
or (day < 1) or (day > 31) then
return
end
temp = div(month - 14, 12)
return (
day - 32075 +
div(1461 * (year + 4800 + temp), 4) +
div(367 * (month - 2 - temp * 12), 12) -
div(3 * div(year + 4900 + temp, 100), 4)
) - 2415021
end
I want to be able to present "today" and "yesterday" for recent dates in my application. I've got a date formatter in use currently to show dates (retrieved from data records) and will keep using this for anything more than a couple of days old. I just really like the way the SMS app in the iPhone shows dates for recent messages and would like to emulate this.
The time-stamps that I have to work with are generated on a server that the phone downloads the data records from. All times are therefore generated at UTC (i.e. GMT) time.
I've been fiddling about with this for a while the solutions I've devised just seem horribly long-winded.
Can anyone suggest how to implement a method that could do this?
Cheers - Steve.
If this is a web app, you might find PrettyDate useful. I made a vb.net implementation that could easily be converted to another language:
Public Function formatDate(ByVal time As DateTime) As String
Dim datediff As TimeSpan = Now.Subtract(time)
Dim days As Integer = datediff.TotalDays
If days < 1 Then
Dim seconds As Integer = datediff.TotalSeconds
Select Case seconds
Case 0 To 60
Return "just now"
Case 61 To 120
Return "1 minute ago"
Case 121 To 3600
Return Math.Floor(seconds / 60) & " minutes ago"
Case 3601 To 7200
Return "1 hour ago"
Case 7201 To 86400
Return Math.Floor(seconds / 3600) & " hours ago"
End Select
ElseIf days < 31 Then
Select Case days
Case 1
Return "yesterday"
Case 2 To 7
Return days & " days ago"
Case Is > 7
Return Math.Ceiling(days / 7) & " weeks ago"
End Select
Else : Return time.ToString("MM/dd/yyyy")
End If
End Function