I'm writing some code to parse RSS feeds but I have trouble with the Abstruse Goose RSS feed.
If you look in that feed, dates are encoded as Mon, 06 Aug 2018 00:00:00 UTC. To me, it looks like RFC 2822.
I tried to parse it using chrono's DateTime::parse_from_rfc2822, but I get ParseError(NotEnough).
let pub_date = entry.pub_date().unwrap().to_owned();
return rfc822_sanitizer::parse_from_rfc2822_with_fallback(&pub_date)
.unwrap_or_else(|e| {
panic!(
"pub_date for item {:?} (value is {:?}) can't be parsed due to error {:?}",
&entry, pub_date, e
)
})
.naive_utc();
Is there something I'm doing wrong? Do I have to hack it some way?
I use rfc822_sanitizer which does a good job at fixing bad writing errors (most of the time). I don't think it impacts the parsing ... but who knows?
The RFC2822 date/time format is very well codified in the RFC as the following format:
date-time = [ day-of-week "," ] date FWS time [CFWS]
day-of-week = ([FWS] day-name) / obs-day-of-week
day-name = "Mon" / "Tue" / "Wed" / "Thu" /
"Fri" / "Sat" / "Sun"
date = day month year
year = 4*DIGIT / obs-year
month = (FWS month-name FWS) / obs-month
month-name = "Jan" / "Feb" / "Mar" / "Apr" /
"May" / "Jun" / "Jul" / "Aug" /
"Sep" / "Oct" / "Nov" / "Dec"
day = ([FWS] 1*2DIGIT) / obs-day
time = time-of-day FWS zone
time-of-day = hour ":" minute [ ":" second ]
hour = 2DIGIT / obs-hour
minute = 2DIGIT / obs-minute
second = 2DIGIT / obs-second
zone = (( "+" / "-" ) 4DIGIT) / obs-zone
Where obs-zone is defined as follows:
obs-zone = "UT" / "GMT" / ; Universal Time
; North American UT
; offsets
"EST" / "EDT" / ; Eastern: - 5/ - 4
"CST" / "CDT" / ; Central: - 6/ - 5
"MST" / "MDT" / ; Mountain: - 7/ - 6
"PST" / "PDT" / ; Pacific: - 8/ - 7
%d65-73 / ; Military zones - "A"
%d75-90 / ; through "I" and "K"
%d97-105 / ; through "Z", both
%d107-122 ; upper and lower case
Something a lot of people get wrong when rolling their own timestamp generation library is this particular point - how to properly label an RFC2822 TZ offset. The reason UT is as it is is because UTC and UT are not exactly the same (one has additional seconds, the other has... four variants! And the RFC does not define which one is used; they're all subtly different).
Related
How to calculate Duration of Time between two date time through VBscript
Date1 = 2021-01-22 11:43:38.000
Date2 = 2021-01-22 14:32:38.000
result should be HH:MM:SS
TimeSerial and FormatDateTime return a date or time that will take the regional settings of the computer into account. On my European computer no AM extension is shown because we use the 24h time format.
An additional problem with TimeSerial is that it will overflow once there are more than 32767 seconds.
A different approach could be to calculate the values for hours, minutes and seconds separately. A possible solution could be:
secValue = DateDiff("s",Date1,Date2)
hours = secValue \ 3600
hh = hours
if hours < 10 then
hh = Right("0" & hours, 2)
end if
mm = Right("0" & (secValue - hours * 3600) \ 60, 2)
ss = Right("0" & secValue mod 60, 2)
diff = hh & ":" & mm & ":" & ss
wscript.echo diff
Finally i answered to my Question Feeling great
Date1 = alA_Filling(0)
Date2 = alA_Filling(1)
secValue = DateDiff("s",Date1,Date2)
ts = TimeSerial(0, 0, secValue)
Duration = FormatDateTime(ts, vbLongTime)
But i got output like 2:49:00 AM why added AM to this may be this vbLongTime
how can i remove that AM
I have dates that look like "17-JAN-19", "18-FEB-20". When I attempt to use the Dates package Date("17-JAN-19", "d-u-yy") I get reasonably 0019-01-17. I could do Date("17-JAN-19", "d-u-yy") + Year(2000) but that introduces the possibility of new errors (I was going to give the example of leap year but that generally works though there is the very rare error Date("29-FEB-00", "d-u-yy")+Year(1900)).
Is there a date format that embeds known information about century?
As mentioned in https://github.com/JuliaLang/julia/issues/30002 there are multiple heuristics for assigning the century to a date. I would recommend being explicit and handling it through a helper function.
const NOCENTURYDF = DateFormat("d-u-y")
"""
parse_date(obj::AbstractString,
breakpoint::Integer = year(now()) - 2000,
century::Integer = 20)
Parses date in according to DateFormat("d-u-y") after attaching century information.
If the year portion is greater that the current year,
it assumes it corresponds to the previous century.
"""
function parse_date(obj::AbstractString,
breakpoint::Integer = year(now()) - 2000,
century::Integer = 20)
# breakpoint = year(now()) - 2000
# century = year(now()) ÷ 100
#assert 0 ≤ breakpoint ≤ 99
yy = rpad(parse(Int, match(r"\d{2}$", obj).match), 2, '0')
Date(string(obj[1:7],
century - (parse(Int, yy) > breakpoint),
yy),
NOCENTURYDF)
end
parse_date("17-JAN-19")
parse_date("29-FEB-00")
I'm having a problem with Coldfusion's DateDiff(). I'm trying to get the difference between two dates with times, like the following examples:
fromdate=06/11/2017 22:10
todate =16/11/2017 23:20
should return:
10 days, 1 hour and 10 minutes
fromdate=06/11/2017 22:10
todate =16/11/2017 20:20
should return:
9 days, 22 hours, 10 minutes
Any help?
Code:
<cfset dtFrom = "11/06/2017 22:10" />
<cfset dtTo = "11/16/2017 23:20" />
<cfoutput>
#DateDiff( "d", dtFrom, dtTo)# Days,
#DateDiff( "h", dtFrom, dtTo) % 24# Hours
#DateDiff( "n", dtFrom, dtTo) % 24 % 60# Minutes
</cfoutput>
In addition to the previous suggestion, DateDiff() isn't going to understand those specific strings or that "06/11/2017" should mean November 6. The result will be:
158 days 1 hours 10 minutes
For it to work as expected, you must convert the strings into date objects first. For example use LSParseDateTime with the right Locale.
fromDate = lsParseDateTime("06/11/2017 22:10", "English (UK)", "dd/MM/yyyy hh:mm");
toDate = lsParseDateTime("16/11/2017 23:20", "English (UK)", "dd/MM/yyyy hh:mm");
or possibly:
fromDate = lsParseDateTime("06/11/2017 22:10", "English (UK)");
toDate = lsParseDateTime("16/11/2017 23:20", "English (UK)");
Here is one way.
totalMinutes = datediff("n", fromDate, toDate);
days = int(totalMinutes /(24 * 60)) ;
minutesRemaining = totalMinutes - (days * 24 * 60);
hours = int(minutesRemaining / 60);
minutes = minutesRemaining mod 60;
writeoutput(days & ' days ' & hours & ' hours ' & minutes & ' minutes');
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 am using a mailing list program which inserts a date into a web link that is "encoded" so it can't be changed or edited by users.
The format is described as follows:
An eight character string, AABBCCDD,where:
Year = 1980 + HexToInt(BB) / 3
Month = HexToInt(CC) / 7 - 21
Day = HexToInt(DD) / 7 - 5
There is also a checksum included to avoid casual modification:
AA = IntToHex(Year + Month + Day mod 200)
For example 2660BDAF would refer to 20 June, 2012.
Can you help me convert the following to Classic ASP:
CodedDateStr = Request.querystring("Exp")
AYear = 1980 + HexToInt(CodedDateStr[3] + CodedDateStr[4]) / 3
AMonth = HexToInt(CodedDateStr[5] + CodedDateStr[6]) / 7 - 21
ADay = HexToInt(CodedDateStr[7] + CodedDateStr[8]) / 7 - 5
ACheckSum = AYear + AMonth + ADay mod 200
if ACheckSum <> HexToInt(CodedDateStr[1] + CodedDateStr[2]) then
ValidDate = 0
else
ValidDate = 1
end if
AExpiryDate = EncodeDate(ADay, AMonth, AYear)
if Date() > AExpiryDate then
ExpiredOffer = 1
else
ExpiredOffer = 0
end if
....
It looks like the HexToInt equivalent is clng("&h" & hexnumber)
I'm not sure about EncodeDate, i hope it is not something cludgy like CDate(AMonth + "/" + ADay + "/" + AYear)
CLng("&h" & hexnumber) looks like a good method for HexToInt.
For EncodeDate, look at the DateSerial function, which takes a year, month, and day, and returns a Date value.