I tried to make a tasks.loop() for checking a muted user that needs to be unmuted, but there's a few problem while doing this, i can't use fetchall() for some reason because it will gives me this error
toremove = muteremove[2]
IndexError: list index out of range
If i use fetchone() maybe it only fetch 1 user every 10 secs, i mean how to fetch all the data every 10 sec to unmute a user?
Also if i use fetchone() it will say that it can't convert str into datetime.datetime object, how can i fix this?
#tasks.loop(seconds=10)
async def muted_user_check(self):
self.cur.execute(f"SELECT userId, guildId, expiredAt FROM mutedlist")
muteremove = self.cur.fetchall()
if muteremove is None:
print("No user to unmute :D")
if muteremove is not None:
toremove = muteremove[2]
timenow = datetime.utcnow()
if timenow > toremove:
self.cur.execute(f"DELETE FROM mutedlist WHERE guildId = {muteremove[1]} and userId = {muteremove[0]}")
To convert a string into a datetime object, you can use the strptime() method:
from datetime import datetime
def convert(date, format):
return datetime.strptime(date, format)
[input] convert('22/08/2020', '%d/%m/%Y')
[output] 2020-08-22 00:00:00
The output will be a datetime object that you can format with the strftime() method like so:
#Example
from datetime import datetime
now = datetime.now() #now will be a datetime object
now.strftime('%d/%m/%Y - %H:%M:%S') # DD/MM/YYYY - hours:minutes:seconds
Here's a list of some formats:
%A → Weekday (%a for abreviations and %w for numbers)
%-d → day of the mount (1, 2, 3, 4, ...)
%B → Mounth name (%b for abreviations and %-m for numbers)
%I → Hour (12h clock)
%p → AM or PM
%H → Hour (24h clock)
%M → Minutes
%S → Seconds
%f → Microseconds
%c → Local date and time representation
Using your code, it would be:
#tasks.loop(seconds=10)
async def muted_user_check(self):
self.cur.execute(f"SELECT * FROM mutedlist")
mute_list = self.cur.fetchall()
if not mute_list:
print("No user to unmute :D")
else:
timeNow = datetime.utcnow()
for mute in mute_list:
muteExpire = datetime.strptime(mute[3], '%Y-%m-%d %H:%M:%S')
if timeNow > muteExpire :
self.cur.execute(f"DELETE FROM mutedlist WHERE guildId=? AND userId=?", (mute[0], mute[1]))
Related
I have a datetime string this format
44340.5416666667 but i want to convert this to 5/24/2021 3:00:00 PM - 4:00:00 PM format. How can i parse that with golang? I tried some convert function but it didn't work.
According to https://kb.paessler.com/en/topic/1313-how-do-i-translate-prtg-timestamp-values-format-to-normal-time-format, the timestamp format used by PRTG seems to be defined as the value of days since Dec 30, 1899.
Following the above link, the following Go code should convert the timestamp into a Go Time instance:
prtg := 44340.5416666667
// substract number of days between Dec 30, 1899 and Jan 1, 1970 and convert to millis
millis := int64((prtg - 25569) * 86400 * 1000)
t := time.Unix(0, millis*int64(time.Millisecond))
println(t.Format("1/2/2006 03:04:05 PM"))
According to prtg timestamp mentioned in Gregor Zurowski's comment,
convert your time to nano seconds (minimum unit in time to more accurate) and add unix nano of 1899-12-30 12.00 midnight.
re convert it to time and format it as below
package main
import (
"fmt"
"time"
)
func main() {
startDate := time.Date(1899, 12, 30, 0, 0, 0, 0, time.UTC).UnixNano()
timeVar := 44340.5416666667 //your time variable
duration := startDate + int64(float64(24*60*60) * timeVar * 1e9) //duration since start date in nanoseconds
fmt.Println(time.Unix(0, duration).Format("1/2/2006 03:04:05 PM"))
}
Noob here,
I'm stuck at trying to present user input in military time into standard time. The code works so far, but I need to subtract 12 hours from the end time to display in standard time. How do I do this using datetime.time? Also, do I need to convert the original user input to an integer to perform datetime.timedelta calculations? Previous questions don't seem to answer my coding questions.
My code is:
def timeconvert():
print "Hello and welcome to Python Payroll 1.0."
print ""
# User input for start time. Variable stored.
start = raw_input("Enter your check-in time in military format (0900): ")
# User input for end time. Variable stored.
end = raw_input("Enter your check-out time in military format (1700): ")
print ""
# ---------------------------------------------------------------------------
# Present user input in standard time format hhmm = hh:mm
# ---------------------------------------------------------------------------
import datetime, time
convert_start = datetime.time(hour=int(start[0:2]), minute=int(start[2:4]))
# need to find a way to subtract 12 from the hour to present end time in standard time
convert_end = datetime.time(hour=int(end[0:2]), minute=int(end[2:4]))
print 'You started at', convert_start.strftime("%H:%M"),'am', 'and ended at', convert_end.strftime("%H:%M"), 'pm'
# ---------------------------------------------------------------------------
# Use timedelta to caculate time worked.
# ---------------------------------------------------------------------------
# print datetime.timedelta
timeconvert()
raw_input("Press ENTER to exit program") # Closes program.
Thanks.
You can use strftime("%I:%M %p") to get standard 12 hour formatting with "AM" or "PM" at the end. See the Python documentation for more details on datetime string formatting.
Also, while it is not natively supported, you can simply use the two datetime.time instances to do your calculation as part of the timedelata constructor.
The below code should suffice, though proper error checking should definitely be used. ;)
--ap
start = raw_input("Enter your check-in time in military format (0900): ")
end = raw_input("Enter your check-out time in military format (1700): ")
# convert user input to datetime instances
start_t = datetime.time(hour=int(start[0:2]), minute=int(start[2:4]))
end_t = datetime.time(hour=int(end[0:2]), minute=int(end[2:4]))
delta_t = datetime.timedelta(
hours = (end_t.hour - start_t.hour),
minutes = (end_t.minute - start_t.minute)
)
# datetime format
fmt = "%I:%M %p"
print 'You started at %s and ended at %s' % (start_t.strftime(fmt), end_t.strftime(fmt))
print 'You worked for %s' % (delta_t)
def time12hr(string):
hours = string[:2]
minutes = string[2:]
x = " "
if int(hours) == 12:
x = "p.m."
hours = "12"
elif int(hours) == 00:
x = "a.m."
hours = "12"
elif int(hours) > 12:
x = "p.m."
hours = str(int(hours) - 12)
else:
x = "a.m."
return "%s:%s %s"%(hours ,minutes,x)
print time12hr('1202')
print time12hr('1200')
print time12hr('0059')
print time12hr('1301')
print time12hr('0000')
I need to standardise and compare date/time fields that are in differnt timezones. eg How do you find the time difference between the following two times?...
"18-05-2012 09:29:41 +0800"
"18-05-2012 09:29:21 +0900"
What's the best way to initialise standard varaibles with the date/time?
The output needs to display the difference and normalised data in a timezone (eg +0100) that is different to the incoming values and different to the local environment.
Expected Output:
18-05-2012 02:29:41 +0100
18-05-2012 01:29:21 +0100
Difference: 01:00:20
import java.text.SimpleDateFormat
def dates = ["18-05-2012 09:29:41 +0800",
"18-05-2012 09:29:21 +0900"].collect{
new SimpleDateFormat("dd-MM-yyyy HH:mm:ss Z").parse(it)
}
def dayDiffFormatter = new SimpleDateFormat("HH:mm:ss")
dayDiffFormatter.setTimeZone(TimeZone.getTimeZone("UTC"))
println dates[0]
println dates[1]
println "Difference "+dayDiffFormatter.format(new Date(dates[0].time-dates[1].time))
wow. doesn't look readable, does it?
Or, use the JodaTime package
#Grab( 'joda-time:joda-time:2.1' )
import org.joda.time.*
import org.joda.time.format.*
String a = "18-05-2012 09:29:41 +0800"
String b = "18-05-2012 09:29:21 +0900"
DateTimeFormatter dtf = DateTimeFormat.forPattern( "dd-MM-yyyy HH:mm:ss Z" );
def start = dtf.parseDateTime( a )
def end = dtf.parseDateTime( b )
assert 1 == Hours.hoursBetween( end, start ).hours
Solution:
Groovy/Java Date objects are stored as the number of milliseconds after
1970 and so do not contain any timezone information directly
Use Date.parse method to initialise the new date to the specified format
Use SimpleDateFormat class to specify the required output format
Use SimpleDateFormat.setTimeZone to specifiy the timezone of the output
data
By using European/London timezone rather than GMT it will
automatically adjusts for day light savings time
See here for a full list of the options for date time patterns
-
import java.text.SimpleDateFormat
import java.text.DateFormat
//Initialise the dates by parsing to the specified format
Date timeDate1 = new Date().parse("dd-MM-yyyy HH:mm:ss Z","18-05-2012 09:29:41 +0800")
Date timeDate2 = new Date().parse("dd-MM-yyyy HH:mm:ss Z","18-05-2012 09:29:21 +0900")
DateFormat yearTimeformatter = new SimpleDateFormat("dd-MM-yyyy HH:mm:ss Z")
DateFormat dayDifferenceFormatter= new SimpleDateFormat("HH:mm:ss") //All times differences will be less than a day
// The output should contain the format in UK time (including day light savings if necessary)
yearTimeformatter.setTimeZone(TimeZone.getTimeZone("Europe/London"))
// Set to UTC. This is to store only the difference so we don't want the formatter making further adjustments
dayDifferenceFormatter.setTimeZone(TimeZone.getTimeZone("UTC"))
// Calculate difference by first converting to the number of milliseconds
msDiff = timeDate1.getTime() - timeDate2.getTime()
Date differenceDate = new Date(msDiff)
println yearTimeformatter.format(timeDate1)
println yearTimeformatter.format(timeDate2)
println "Difference " + dayDifferenceFormatter.format(differenceDate)
I have the following f# code:
let mutable argNum = 0
let cmdArgs = System.Environment.GetCommandLineArgs()
for arg in cmdArgs do
printfn "arg %d : %s" argNum arg
match argNum with
| 1 -> pmID <- System.Int32.Parse arg
| 2 -> startDate <- System.DateTime.Parse arg
| 3 -> endDate <- System.DateTime.Parse arg
| _ -> ()
argNum <- argNum + 1
for the date parameters, the argument comes in the form: "1-1-2011", so "M-D-YYYY"
when i write the dates into the xml serializer, I get the following format:
1/1/2011 12:00:00 AM
I'd like to remove the Time piece completely. What's the best way to do this?
What about this one?
(DateTime.Parse arg).ToString("MM-dd-yyyy")
By default DateTime instances are serialized using the "dateTime" datatype for serialization. You can change it by annotating your type to use "date" instead, which will serialize it in the format YYYY-MM-DD - if this doesn't work for you just serialize a string instead that holds the format of your choice.
You can set the custom serialization attribute like this:
full datetime: [<XmlAttribute("start-date", DataType = "dateTime")>]
just the date part: [<XmlAttribute("start-date", DataType = "date")>]
Example:
[<Serializable>]
type DateTest() =
let mutable startDate = DateTime.Now
[<XmlAttribute("start-date", DataType = "date")>]
member x.StartDate with get() = startDate and set v = startDate <- v
I have a tab delimited file where each record has a timestamp field in 12-hour format:
mm/dd/yyyy hh:mm:ss [AM|PM].
I need to quickly convert these fields to 24-hour time:
mm/dd/yyyy HH:mm:ss.
What would be the best way to do this? I'm running on a Windows platform, but I have access to sed, awk, perl, python, and tcl in addition to the usual Windows tools.
Using Perl and hand-crafted regexes instead of facilities like strptime:
#!/bin/perl -w
while (<>)
{
# for date times that don't use leading zeroes, use this regex instead:
# (?:\d{1,2}/\d{1,2}/\d{4} )(\d{1,2})(?::\d\d:\d\d) (AM|PM)
while (m%(?:\d\d/\d\d/\d{4} )(\d\d)(?::\d\d:\d\d) (AM|PM)%)
{
my $hh = $1;
$hh -= 12 if ($2 eq 'AM' && $hh == 12);
$hh += 12 if ($2 eq 'PM' && $hh != 12);
$hh = sprintf "%02d", $hh;
# for date times that don't use leading zeroes, use this regex instead:
# (\d{1,2}/\d{1,2}/\d{4} )(\d{1,2})(:\d\d:\d\d) (?:AM|PM)
s%(\d\d/\d\d/\d{4} )(\d\d)(:\d\d:\d\d) (?:AM|PM)%$1$hh$3%;
}
print;
}
That's very fussy - but also converts possibly multiple timestamps per line.
Note that the transformation for AM/PM to 24-hour is not trivial.
12:01 AM --> 00:01
12:01 PM --> 12:01
01:30 AM --> 01:30
01:30 PM --> 13:30
Now tested:
perl ampm-24hr.pl <<!
12/24/2005 12:01:00 AM
09/22/1999 12:00:00 PM
12/12/2005 01:15:00 PM
01/01/2009 01:56:45 AM
12/30/2009 10:00:00 PM
12/30/2009 10:00:00 AM
!
12/24/2005 00:01:00
09/22/1999 12:00:00
12/12/2005 13:15:00
01/01/2009 01:56:45
12/30/2009 22:00:00
12/30/2009 10:00:00
Added:
In What is a Simple Way to Convert Between an AM/PM Time and 24 hour Time in JavaScript, an alternative algorithm is provided for the conversion:
$hh = ($1 % 12) + (($2 eq 'AM') ? 0 : 12);
Just one test...probably neater.
It is a 1-line thing in python:
time.strftime('%H:%M:%S', time.strptime(x, '%I:%M %p'))
Example:
>>> time.strftime('%H:%M:%S', time.strptime('08:01 AM', '%I:%M %p'))
'08:01:00'
>>> time.strftime('%H:%M:%S', time.strptime('12:01 AM', '%I:%M %p'))
'00:01:00'
Use Pythons datetime module someway like this:
import datetime
infile = open('input.txt')
outfile = open('output.txt', 'w')
for line in infile.readlines():
d = datetime.strptime(line, "input format string")
outfile.write(d.strftime("output format string")
Untested code with no error checking. Also it reads the entire input file in memory before starting.
(I know there is plenty of room for improvements like with statement...I make this a community wiki entry if anyone likes to add something)
To just convert the hour field, in python:
def to12(hour24):
return (hour24 % 12) if (hour24 % 12) > 0 else 12
def IsPM(hour24):
return hour24 > 11
def to24(hour12, isPm):
return (hour12 % 12) + (12 if isPm else 0)
def IsPmString(pm):
return "PM" if pm else "AM"
def TestTo12():
for x in range(24):
print x, to12(x), IsPmString(IsPM(x))
def TestTo24():
for pm in [False, True]:
print 12, IsPmString(pm), to24(12, pm)
for x in range(1, 12):
print x, IsPmString(pm), to24(x, pm)
This might be too simple thinking, but why not import it into excel, select the entire column and change the date format, then re-export as a tab delimited file? (I didn't test this, but it somehow sounds logical to me :)
Here i have converted 24 Hour system to 12 Hour system.
Try to use this method for your problem.
DateFormat fmt = new SimpleDateFormat("yyyyMMddHHssmm");
try {
Date date =fmt.parse("20090310232344");
System.out.println(date.toString());
fmt = new SimpleDateFormat("dd-MMMM-yyyy hh:mm:ss a ");
String dateInString = fmt.format(date);
System.out.println(dateInString);
} catch (Exception e) {
System.out.println(e.getMessage());
}
RESULT:
Tue Mar 10 23:44:23 IST 2009
10-March-2009 11:44:23 PM
In Python: Converting 12hr time to 24hr time
import re
time1=input().strip().split(':')
m=re.search('(..)(..)',time1[2])
sec=m.group(1)
tz=m.group(2)
if(tz='PM'):
time[0]=int(time1[0])+12
if(time1[0]=24):
time1[0]-=12
time[2]=sec
else:
if(int(time1[0])=12):
time1[0]-=12
time[2]=sec
print(time1[0]+':'+time1[1]+':'+time1[2])
Since you have multiple languages, I'll suggest the following algorithm.
1 Check the timestamp for the existence of the "PM" string.
2a If PM does not exist, simply convert the timestamp to the datetime object and proceed.
2b If PM does exist, convert the timestamp to the datetime object, add 12 hours, and proceed.