I'm stumped as to why the following PHP strtotime function returns '07' as the month number, rather than '06' when $monthToGet = 'June':
$monthToGet = $_GET['mon'];
$monthAsNumber = date('m', strtotime($monthToGet));
From searching, it appears it may be due to default date parameters (in this case the day and year) as I haven't specified them. Would that be the cause?
Any suggestions appreciated!
TL;DR
You are right
echo date("m", strtotime("June"));
-> 07
However, this does work:
echo date("m", strtotime("1. June 2012"));
-> 06
The problem explained
Today is 31. July 2012 and since you provide only a month, the current day and current year are used to create a valid date.
See the documentation:
NOTE
The function expects to be given a string containing an English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
Alternatives
You could use date_parse_from_format() or strptime() to achieve what you want with a slightly different approach.
(Thanks to johannes_ and johann__ for their input)
Fixed with :
$monthToGet = '1 '. $_GET['mon'];
But I still don't get why, since "m" is a valid date format
Today is 31 Jul. So a strtotime with only "June" is interpreted as 31 June => 1 July.
In fact:
echo date("Y-m-d",strtotime("January")); // 2012-01-31
echo date("Y-m-d",strtotime("February")); // 2012-03-02
of course... only today 31 Jul 2012 :) Tomorrow all will works.
You're lucky because you found this bug just today ;)
Related
Code:
$time = strtotime('2020-03-31');
echo date('Y-m-d', strtotime('-1 month', $time));
Expected Result: Any date from Feb 2020
Actual Result: 2020-03-02
Is there any better way to add or subtract a month from a given date?
Months are an awkward interval to work with, because they don't have a fixed length. Should the algorithm assume that by "1 month" you mean "30 days", or "31 days", or should it just try subtracting 1 from the "month" field in the date structure?
The last option is what is happening here: given "2020-03-31", PHP's date library is subtracting 1 from the "03" to give "2020-02-31". Since that's an invalid date (February 2020 had 29 days), it then "normalises" it to a real date - 2 days after the 29th February was the 2nd March.
Probably you want to use a more specific period to subtract, like 30 days - although note that if the initial input is "2020-03-01" that will give you "2020-01-31", not "2020-02-01".
Ultimately, this is a problem with our irregular calendar, rather than with PHP. It's really up to you to define what you mean by "a month before", and use a more specific algorithm that captures that requirement.
You can make code like below
<?php
$time = strtotime('2020-03-1 -32 days');
echo date('M-Y', $time); // output Feb-2020
?>
The above code will return date as you expected
i am trying to get the day of week from a timestamp:
an example of the timestamp could be:
2014-09-14 18:28:11
I have tried with the following code:
$date = date("D", strtotime($activity[$i]['timestamp']));
However the result i get here is:
Thu
which should have been sunday?
Also is it possible to get it as a full discription instead of a short version of the day name?
Answer to part two of the question is that you can just use l (lowercase 'L') and it'll output Sunday instead of Sun.
$date = date("l", strtotime($activity[$i]['timestamp']));
As for the first part, it probably output Thursday, 1 January 1970 because it received an error instead of an actual date as argument to strtotime.
I know this is a common problem, but I can't seem to find the solution anywhere. In my last question [complicated date functions - comparing, subtracting I needed to compare timestamps to get an accurate date for some Cisco logs.
The best I can come up with (since the dates don't actually feature the year) is in the format
Mar 1 2013 00:03:55:
from
Mar 1 00:03:55:
But when I ran some tests, strtotime is converting this date as
Jan 1970
Using
print date("M Y", strtotime($c_log))."\n";
Am I going to have to reformat it into a date it can understand? I don't appear to have the DateTime function. What's the simplest way?
Use the date_parse_from_format() function so you can specify the format
http://php.net/manual/en/function.date-parse-from-format.php
Works for me:
echo date('M Y', strtotime('Mar 1 00:03:55')); // outputs Mar 2013
it should default to the current year when one isn't included.
I'm trying to convert a date to a string by using the toString('dd MMMM YYY') function.
But here, this isn't working.
Here is my code:
$date_start = new Zend_Date(strtotime($this->startdate));
echo($date_start);
The result is: 31 Dec 2012 00:00:00
$date_input = $date_start->toString('dd MMMM YYY');
echo($date_input);
The result is: 31 December 2013
What do I have to do to obtain 31 December 2012 ?
This is a known issue described here: http://framework.zend.com/issues/browse/ZF-5297
Note that the default ISO format differs from PHP's format which can be irritating if you have not used in previous. Especially the format specifiers for Year and Minute are often not used in the intended way.
For year there are two specifiers available which are often mistaken. The Y specifier for the ISO year and the y specifier for the real year. The difference is small but significant. Y calculates the ISO year, which is often used for calendar formats. See for example the 31. December 2007. The real year is 2007, but it is the first day of the first week in the week 1 of the year 2008. So, if you are using 'dd.MM.yyyy' you will get '31.December.2007' but if you use 'dd.MM.YYYY' you will get '31.December.2008'. As you see this is no bug but a expected behaviour depending on the used specifiers.
Use "yyy" instead of "YYY".
Use yyy instead of YYY.
YYY is the ISO-8601 date, which is different from a calendar date.
http://framework.zend.com/manual/1.12/en/zend.date.constants.html
I am having a problem with php's DateTime functions.
Today is monday 3 december.
Assuming the following code:
$dte = new DateTime(date('Y-m-d H:i:s'));
var_dump($dte->format('Y-W'));
$dte->modify('+4 weeks');
var_dump($dte->format('Y-m-d H:i:s -- Y_W'));
$dte->modify('+1 days');
var_dump($dte->format('Y-m-d H:i:s -- Y_W'));
After four weeks it would be 31st of december. I would suspect to get the last week of the year (52?). But what I get is week 1 of 2012 as you can see in the following output.
string '2012-49' (length=7)
string '2012-12-31 14:48:00 -- 2012_01' (length=30)
string '2013-01-01 14:48:00 -- 2013_01' (length=30)
So my problem is that after the first modification I think I should get:
2012-12-31 14:48:00 -- 2012_52
but instead I get
2012-12-31 14:48:00 -- 2012_01
So why does the week go back to 01 without incrementing the year, and than why does the other line gives me 2013_01 ?
EDIT::
I now see that the week before is week 52, anything to do with leap year?
But then again, how can the week go back to 01 without incrementing the year?
So why does the week go back to 01 without incrementing the year, and than why does the other line gives me 2013_01 ?
I think you're displaying the "year" instead of the "week-year". When you're using week numbers, it's the week-year that's the relevant part; simple "year" is only relevant with respect to month and day.
EDIT: I think you want the o format specifier instead, so try:
var_dump($dte->format('Y-m-d H:i:s -- o_W'));
That should show you 2013_01 for December 31st 2012, as it's in week 1 of week-year 2013.
So basically, I don't think this is a bug in DateTime - it's just a misunderstanding of how "week of year" is meant to be used.
This seems to be no bug. According to the documentation W will return ISO-8601 week number of year, weeks starting on Monday (added in PHP 4.1.0) and because Mon, 31 Dec 2012 15:04:46 +0100 is Monday it will be 1 instead of 52.
Further information on Wikipedia and this nice site.