How to format a full year date/time

How to format a full year date/time

By : user3045453
Date : December 02 2020, 10:46 PM
I wish this helpful for you The Date object isn't "already formatted" - when you pass it in this way its toString() method is invoked and returned as an ECMA-262-compliant string representation based on the information associated with tha instance of Date.
You can cobble your format together using a combination of the Date object's toLocaleString(), getDate() and getFullYear() methods, like so:
code :
var today = new Date();
var quarter = Math.floor((today.getMonth() + 3) / 3);
var nextq;
if (quarter == 4) {
    nextq = new Date (today.getFullYear() + 1, 1, 1);
} else {
    nextq = new Date (today.getFullYear(), quarter * 3, 1);

console.log("nextq: " + nextq.toLocaleString('en-us', {  weekday: 'long' }) + ", " + nextq.toLocaleString('default', { month: 'long' }) + " " + ("0" + nextq.getDate()).slice(-2) + ", " + nextq.getFullYear());

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Convert DDMON (eg. 23JAN) date format to full date with correct year applied in Ruby

Convert DDMON (eg. 23JAN) date format to full date with correct year applied in Ruby

By : Alejandro Cabrera
Date : March 29 2020, 07:55 AM
I hope this helps you . Am looking for a compact solution for converting the dates that come from the airline reservation systems (GDS like Sabre/Galileo/Amadeus) which dont have the year - eg. 23JUN to a standard date, but need to determine the year quickly to tack on in Ruby. (FYI - on these systems date can only be max 355 days from today. So there is no ambiguity of interpreting 12MAY as 2013-05-12 instead of 2012-05-12 assuming today is 12MAY 2012). If today is 25th Dec 2012, and date is entered as 01JAN, this would be 01JAN2013 in the future, and if today is 21JUL 2012, input date of 01JAN would also become 01JAN 2013 as it is a past date in the current year. , How about this?
code :
t = DateTime.now
crsdate = DateTime.strptime("23JAN","%d%b")
crsdate >= t ? crsdate >>= 12 : crsdate
Format a Date Time to the Full date/time pattern (short time) in Orchard

Format a Date Time to the Full date/time pattern (short time) in Orchard

By : Lars Hjelkrem
Date : March 29 2020, 07:55 AM
like below fixes the issue {Content.Fields.MyContentType.MyDateTimeField.DateTime.Local.Format:dddd, MMMM d, yyyy h:mm tt} should do the trick.
How to format date with a IntlDateFormatter using MEDIUM or FULL datetype, but without a year?

How to format date with a IntlDateFormatter using MEDIUM or FULL datetype, but without a year?

By : Arsalan Muhmmad Huss
Date : March 29 2020, 07:55 AM
hop of those help? So, currently it's impossible. I've found that intl PHP extension simply lacks of DateTimePatternGenerator (http://userguide.icu-project.org/formatparse/datetime), which is exactly what I need.
code :
locale | format pattern for skeleton “MMMMdjmm” | example
en_US    "MMMM d  'at'  h:mm a"                   April 2 at 5:00 PM
es_ES    "d 'de' MMMM, H:mm"                      2 de abril, 17:00
ja_JP    "M月d日 H:mm"                              4月2日 17:00
Converting 4-digit year format to Python MM/DD/YYYY date time format

Converting 4-digit year format to Python MM/DD/YYYY date time format

By : Piyush Pahwa
Date : March 29 2020, 07:55 AM
seems to work fine I used regex .findall(' ') function to extract date time elements of different formats from strings within a pandas data frame to 'list' objects and made them into a new column named "new." However, there are date time list objects that only of 4-digit YYYY format, lacking month (MM) and day (DD) (e.g., df['new'].iloc[99]), and objects where day (DD) are missing, such as df['new'].iloc[221], like the following:
code :
import pandas as pd
import re
from datetime import datetime 

def helper(a ,f):
    return datetime.strptime(a,f).strftime('%m-%d-%Y')

def change_format(a):
    #print a
    if 'Janaury' in a:
        a = a[:3]+a[7:]
    if 'Decemeber' in a:
        a = a[:3]+a[9:]
    c = re.split('/|-| ',a)
    b = len(c)
    if re.match(r'\d\d [A-Z]',a) != None:
        if len(c[1]) == 3:
            return helper(a,'%d %b %Y')
            return helper(a, '%d %B %Y')
    elif re.match(r'[A-Z]',a) != None:
        if len(c) == 2:
            if len(c[0]) == 3: 
                return helper(a+' 1','%b %Y %d')
                return helper(a+' 1','%B %Y %d')
        if len(c[0]) == 3:
            if ',' in a:
                return helper(a,'%b %d, %Y')
                return  helper(a,'%b %d %Y')
            if ',' in a:
                return helper(a,'%B %d, %Y')
                return helper(a,'%B %d %Y')
        if b==3:
            if len(c[2]) == 2:
                if '-' in a:
                    return helper(a,'%m-%d-%y')
                    return helper(a ,'%m/%d/%y')
            elif len(c[2]) == 4:
                return date(int(c[2]),int(c[0]),int(c[1])).strftime('%m-%d-%Y')
        elif b==2:
            return date(int(c[1]),int(c[0]),1).strftime('%m-%d-%Y')
            return date(int(c[0]),1,1).strftime('%m-%d-%Y')

with open('dates.txt') as f:
    d = f.read()
k = re.findall(r'\b(\d{1,2}\/\d{1,2}\/\d{2,4}|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*-\d{2}-\d{2,4}|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{2}. )\d{4}|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{2}, )\d{4}|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{2} )\d{4}|(?:\d{2} )(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{4})|(?:\d{2} )(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z.]* (?:\d{4})|(?:\d{2} )(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z,]* (?:\d{4})|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{1,2})[a-z,]* (?:\d{4})|(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]* (?:\d{4})|\d{1,2}\/\d{4}|\d{4}|\d{1,2}-\d{1,2}-\d{2,4})\b',d)   
dates =map(change_format,k)
df = pd.DataFrame(dates,columns= ['date'])
df['date'] =pd.to_datetime(df.date)
df = df.sort_values('date').reset_index(drop=True)
Convert full format date to year month format in Python

Convert full format date to year month format in Python

By : iker
Date : March 29 2020, 07:55 AM
should help you out My dataframe have one column date which has the following data: , It works:
code :
df['date'] = pd.to_datetime(df['date']).dt.to_period('M')
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