Laravel create date from string - datetime

I have a calendar that posts the value in the following format
3 January, 2017
and when I convert it to carbon by doing
$carbon = Carbon::parse($data['due_date']);
echo $carbon;
I see 2016-01-03 20:17:00
My expected output is 2017-01-03

Try
$date = Carbon::createFromFormat('j F, Y', $data['due_date']);

Just use this 👇
echo date('Y-m-d', strtotime('3 January, 2017'));

Related

How to get date of two month old date from current date

I am trying to pull data (some transaction related data) from DB. To pull the data, I am passing start date and end date as an argument to the query.
Here I need to pull the data of last 2 months. i.e., Start time would be Jun 01, 2022 and End time would be Aug 01, 2022.
Below is the script:
#!/usr/bin/perl
use strict;
use warnings;
use DateTime;
use Date::Format;
use Date::Parse;
my $nowdate = DateTime->now(time_zone => 'local');
my ($month, $year) = ($nowdate->month, $nowdate->year);
my $date = DateTime->new(
year => $year,
month => $month,
day => 1,
);
my $end_time = str2time($date);
print "END:$end_time\n";
print time2str("%d-%m-%Y %T", $end_time)."\n"; #printing just to see in human readable format
my $start_time = $date->clone;
$start_time->add( months => 1 )->subtract( days => 92 );
$start_time = str2time($start_time);
print "START:$start_time\n";
print time2str("%d-%m-%Y %T", $start_time)."\n"; #printing just to see in human readable format
I have two issues over here:
I am using DateTime object two times to get end time. Can it be done in one shot?
$start_time->add( months => 1 )->subtract( days => 92 ); In this line of code, I have to explicitly mention subtract 92 days, which wouldn't be right always. Since some months have 30 days or 31 days or even 29 days. How can I get 2 month's beginning day date?
Another example: Lets assume if we are in September 2022, then Start time and End time would be Jul 01, 2022 and Sep 01, 2022 respectively.
Note: Perl version is 5.16.3
It would be good If I can do it with Core modules which comes with 5.16.3
You could simplify it by using truncate(to => 'month') to get to the first in the current month:
my $end = DateTime->now(time_zone => 'local')
->truncate(to => 'month');
This may however fail on a day without a midnight so this may be an option:
my $end = DateTime->now(time_zone => 'local')
->set_time_zone('floating')
->truncate(to => 'month');
Then subtract the number of months to get the start date:
my $start = $end->clone->subtract( months => 2 );
Then:
my $start_time = str2time($start);
my $end_time = str2time($end);
print "START:$start_time\n";
print time2str("%d-%m-%Y %T", $start_time)."\n";
print "END:$end_time\n";
print time2str("%d-%m-%Y %T", $end_time)."\n";
Possible output:
START:1654034400
01-06-2022 00:00:00
END:1659304800
01-08-2022 00:00:00

Convert string to timestamp in symfony 4

I have a variable recover from ajax and in the controller
dd($gethour);
"Sat Oct 26 2019 00:00:04 GMT+0200 (heure d’été d’Europe centrale)"
$from = \DateTime::createFromFormat('m-d-y H:i:s', $gethour);
dd($form);
Result:false
My problem is how to convert string to timestamp seen in the base HoursPass is a type timestamp
It would be better if you could start with a real timestamp which is long number. But otherwise you could do it this way:
$gethour = "Sat Oct 26 2019 00:00:04 GMT+0200 (heure d’été d’Europe centrale)";
$chunks = explode(' ', $gethour);
$from = new \DateTime($chunks[1] . ' ' . $chunks[2] . ' ' . $chunks[3] . ' ' . $chunks[4]);
$from->setTimezone(new DateTimeZone($chunks[5]));
echo $from->format('d-m-Y H:i:s P'); // 26-10-2019 09:00:04 +02:00
you can exam this way :
date("m-d-y H:i:s", strtotime($gethour));
this way is answered for me.

Moment JS getting Date without time [duplicate]

formatCalendarDate = function (dateTime) {
return moment.utc(dateTime).format('LLL');
};
It displays: "28 februari 2013 09:24"
But I would like to remove the time at the end. How can I do that?
I'm using Moment.js.
Sorry to jump in so late, but if you want to remove the time portion of a moment() rather than formatting it, then the code is:
.startOf('day')
Ref: http://momentjs.com/docs/#/manipulating/start-of/
Use format('LL')
Depending on what you're trying to do with it, format('LL') could do the trick. It produces something like this:
Moment().format('LL'); // => April 29, 2016
The correct way would be to specify the input as per your requirement which will give you more flexibility.
The present definition includes the following
LTS : 'h:mm:ss A',
LT : 'h:mm A',
L : 'MM/DD/YYYY',
LL : 'MMMM D, YYYY',
LLL : 'MMMM D, YYYY h:mm A',
LLLL : 'dddd, MMMM D, YYYY h:mm A'
You can use any of these or change the input passed into moment().format().
For example, for your case you can pass moment.utc(dateTime).format('MMMM D, YYYY').
Okay, so I know I'm way late to the party. Like 6 years late but this was something I needed to figure out and have it formatted YYYY-MM-DD.
moment().format(moment.HTML5_FMT.DATE); // 2019-11-08
You can also pass in a parameter like, 2019-11-08T17:44:56.144.
moment("2019-11-08T17:44:56.144").format(moment.HTML5_FMT.DATE); // 2019-11-08
https://momentjs.com/docs/#/parsing/special-formats/
You can also use this format:
moment().format('ddd, ll'); // Wed, Jan 4, 2017
formatCalendarDate = function (dateTime) {
return moment.utc(dateTime).format('LL')
}
Look at these Examples.
Format Dates
moment().format('MMMM Do YYYY, h:mm:ss a'); // December 7th 2020, 9:58:18 am
moment().format('dddd'); // Monday
moment().format("MMM Do YY"); // Dec 7th 20
moment().format('YYYY [escaped] YYYY'); // 2020 escaped 2020
moment().format(); // 2020-12-07T09:58:18+05:30
Relative Time
moment("20111031", "YYYYMMDD").fromNow(); // 9 years ago
moment("20120620", "YYYYMMDD").fromNow(); // 8 years ago
moment().startOf('day').fromNow(); // 10 hours ago
moment().endOf('day').fromNow(); // in 14 hours
moment().startOf('hour').fromNow(); // an hour ago
Calendar Time
moment().subtract(10, 'days').calendar(); // 11/27/2020
moment().subtract(6, 'days').calendar(); // Last Tuesday at 9:58 AM
moment().subtract(3, 'days').calendar(); // Last Friday at 9:58 AM
moment().subtract(1, 'days').calendar(); // Yesterday at 9:58 AM
moment().calendar(); // Today at 9:58 AM
moment().add(1, 'days').calendar(); // Tomorrow at 9:58 AM
moment().add(3, 'days').calendar(); // Thursday at 9:58 AM
moment().add(10, 'days').calendar(); // 12/17/2020
Multiple Locale Support
moment.locale(); // en
moment().format('LT'); // 9:58 AM
moment().format('LTS'); // 9:58:18 AM
moment().format('L'); // 12/07/2020
moment().format('l'); // 12/7/2020
moment().format('LL'); // December 7, 2020
moment().format('ll'); // Dec 7, 2020
moment().format('LLL'); // December 7, 2020 9:58 AM
moment().format('lll'); // Dec 7, 2020 9:58 AM
moment().format('LLLL'); // Monday, December 7, 2020 9:58 AM
moment().format('llll'); // Mon, Dec 7, 2020 9:58 AM
Whenever I use the moment.js library I specify the desired format this way:
moment(<your Date goes here>).format("DD-MMM-YYYY")
or
moment(<your Date goes here>).format("DD/MMM/YYYY")
... etc I hope you get the idea
Inside the format function, you put the desired format. The example above will get rid of all unwanted elements from the date such as minutes and seconds
With newer versions of moment.js you can also do this:
var dateTime = moment();
var dateValue = moment({
year: dateTime.year(),
month: dateTime.month(),
day: dateTime.date()
});
See: http://momentjs.com/docs/#/parsing/object/.
You can use this constructor
moment({h:0, m:0, s:0, ms:0})
http://momentjs.com/docs/#/parsing/object/
console.log( moment().format('YYYY-MM-DD HH:mm:ss') )
console.log( moment({h:0, m:0, s:0, ms:0}).format('YYYY-MM-DD HH:mm:ss') )
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.22.2/moment.min.js"></script>
For people like me want the long date format (LLLL) but without the time of day, there's a GitHub issue for that: https://github.com/moment/moment/issues/2505. For now, there's a workaround:
var localeData = moment.localeData( moment.locale() ),
llll = localeData.longDateFormat( 'llll' ),
lll = localeData.longDateFormat( 'lll' ),
ll = localeData.longDateFormat( 'll' ),
longDateFormat = llll.replace( lll.replace( ll, '' ), '' );
var formattedDate = myMoment.format(longDateFormat);
Try this:
moment.format().split("T")[0]
The thing is - you can run into an issue with timezones. For example, if you parse date like this: '2022-02-26T00:36:21+01:00' it may turn into '25/02/2022' As a solution if your date is in ISO format you can just cut off the time portion from the string, like this:
moment('2022-02-26T00:36:21+01:00'.split('T')[0]).utc().format('DD/MM/YYYY')
This solution is quite blunt, so be careful with string format.
This format works pretty fine
const date = new Date();
const myFormat= 'YYYY-MM-DD';
const myDate = moment(date, 'YYYYMMDDTHHmmss').format(myFormat);
Try
new Date().toDateString()
Result - "Fri Jun 17 2022"
This worked perfectly for me:
moment().format('YYYY-MM-DD')
moment(date).format(DateFormat)
Here DateFormat should be DateFormat = 'YYYY-MM-DD'

Shortest way to convert PayPal's prefered date format to MySQL datetime format in ColdFusion CF9

PayPal's IPN gives dates in this format 06:52:15 Apr 12, 2014 PDT which is not a valid recognised format. Also, the timezone is specified; I can't find a ColdFusion date class that can take a timezone. So, what's the shortest way to convert the format and translate the timezone to GMT (the timezone of my server). So far I have this...
arr3LA = ['Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dev'];
arrDateParts = ListToArray( FORM.payment_date, ' ');
arrTimeParts = ListToArray( arrDateParts[1], ':' );
arrDateParts[3] = Replace(arrDateParts[3], ',', '', 'All');
objPayPalDateTime = CreateDateTime( arrDateParts[4], ArrayFind( arr3LA, arrDateParts[2]), arrDateParts[3], arrTimeParts[1], arrTimeParts[2], arrTimeParts[3] );
strLSFriendly = DateFormat( objPayPalDateTime, 'mmmm dd, yyyy' ) & ' ' & TimeFormat( objPayPalDateTime, 'h:mm:ss tt' ) & ' ' & arrDateParts[5];
objDateTime = LSParseDateTime( strLSFriendly );
strSQLFriendlyDateTime = DateFormat( objDateTime, 'yyyy-mm-dd hh:mm:ss' );
Can you improve on that?
I'd say forget using native CFML stuff here, and leverage java:
target = "06:52:15 Apr 12, 2014 PDT";
sdf = createObject("java", "java.text.SimpleDateFormat").init("hh:mm:ss MMM dd, yyyy zzz");
result = sdf.parse(target);
writeDump([{target=target},{result=result}]);
Does that do the trick?
(NB: inspired by this question: How to parse date string to Date?)

Convert 12-hour date/time to 24-hour date/time

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.

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