Hallo all.
if i have week number 42 and i need to know what dates there are in this week, can sombardy maby tell me a easy way to get this info?
tanks a lot :)
<?php
$date = new DateTime('2009W52');
echo $date->format(DateTime::RFC850);
Which outputs
Monday, 21-Dec-09 00:00:00 EET
and you can modify it like
$date->modify("last monday");
$date->modify("next monday");
Week has to be zero padded for weeks 1-9
There is a thread discussing how to get the first date of the week (Monday) using the week of the year and the year. That should help you.
I know you have already accepted an answer, but I feel this bit of code is also useful.
$year = date("Y");
$week = date("W"); // Can be replaced with '42' for your example.
$start = strtotime($year.'W'.$week.'1');
This will return a unix timestamp which some people find easier to manipulate.
You can also use this for PHP 5.1 and higher.
$start = date(datetime::ISO8601, strtotime("2009W421"));
Using the above method you can easily format it.
Hope this provides useful to someone.
-Mathew
Related
Currently, I'm trying to parse out dates when messages were received into timestamps. I have the month and day but the year is not specified. The event always occurs at the most recent (human) reading of the time. It works great in most cases to do this:
$time = strtotime("Jan 2 8:38pm");
That returns a date for this year, which is correct. Unfortunately, I get problems when I try to do for example:
$time = strtotime("Dec 31 8:38pm");
That returns a date which hasn't happened yet, and wont happen for the whole rest of the year. Obviously, my message was not sent in the future. I need it to return December 31st of last year.
For weekdays, I had a solution by prepending 'last' before the weekday like so:
$time = strtotime("Last Saturday 8:38pm");
That always returned the time of the last Saturday. However, trying to do the same thing here doesn't work:
$time = strtotime("Last Dec 31 8:38pm");
This returns false. I know to decrement a date by 1 year, I can do this:
$time = strtotime("Dec 31 8:38pm -1 year");
And that works great for Dec 31. However, Jan 2 will now fail:
$time = strtotime("Jan 2 8:38pm -1 year");
One solution I thought of was to subtract off a year (86400 * 365) from the resulting value if it is past today's date. However, this result will fail if we passed over February of a leap year. In that case, we would end up with a time that was ahead by a day.
The best solution I came up with so far is this:
$time = strtotime($raw_time);
if ($time > time()) {
$time = strtotime($raw_time." -1 year");
}
It seems kind of wasteful to make two calls to strtotime which I know is probably not a very efficient function. Is this the most elegant solution?
Is anyone aware of an option in strtotime which forces the dates to be in the past instead of in the future?
Is there another way to parse these dates that I should consider?
Efficiency is important for this because I am going to be parsing a lot of dates with it, but I would also like simple and readable code so I can understand it later.
Your approach is fine, as there is no date format to get what you want. Another approach could be using the DateTime class:
$datetime = new DateTime($raw_time);
if ($datetime > new DateTime()) {
$datetime->modify('-1 year');
}
You could test which one of the two approaches is faster. My guess is that this is a micro-optimization that won't make a lot of difference.
I am constructing a diary that has weekly views which I thought I had cracked because the dates seemed to appears as correct. It wasn't until my MySQL queries kept returning what seemed like random results that I realized the the month is actually being seen as the day instead.
$week_number = date("W");
$year = date("Y");
for ($day=0; $day<=6; $day++) {
$daily_date = date('d/m/Y', strtotime($year."W".$week_number.$day))."\n";
$StartDate = date('d', strtotime($daily_date));
}
echo $starteDate;
$startDate returns the number of the month rather than the day and sure enough date('m', strtotime($daily_date)) returns the day rather than the month.
I can't understand where I have made this silly mistake so any help would be appreciated.
This is because of the Americanisation of dates - strtotime will read the date as m/d/Y rather than d/m/Y.
The ISO for dates is Y-m-d and for ease I would use this format when doing any kind of date manipulation.
That code is horrible. You're converting dates to strings multiple times. There is absolutely NO reason to take your $year . w. $week_number.etc... value, convert it to a string, then convert that string back to a date just to extract the day value.
As well, d/m/Y is a horrible format to use for date transport, because... riddle me this, what is 01/02/03. Is that 3rd Feb 2001? 1 Mar 2002? If you can't figure it out, how can you expect strtotime to be better at it? it's fairly smart, but it's not omnicient, and it's DEFINITELY not infallible. A 4digit year does make it a bit easier, but you can still end up with d/m v.s. m/d confusion.
Why not simply
$StartDate = date('d', strtotime($year."W".$week_number.$day));
or better yet, use the DateTime class and select an appropriate DateInterval
I'm trying to get my head round someone else's code which they've written for handling the dates of when news stories are published. The problem has come up because they are using this line -
$date = strtotime("midnight", strtotime($dateString));
to process a date selected using a jquery calendar widget. This works fine for future dates, but when you try to use a date which is in the previous calendar year, it uses the current year instead. I think this is due to "midnight" finding the closest instance of the selected day and month.
I could remove the "midnight", but I'm not sure what the repercussions of this would be - is there a reason that the midnight could be there?
EDIT: this is the full block of code which handles the date. The date contains the time, which allows the user to publish an item at a specific time.
$array['display_date'] = '24 October, 2011 17:30';
$string = $array['display_date'];
$dateString = substr($string, 0, -5);
$timeArray = explode(':', substr($string, -5));
$hours_in_secs = 60 * 60 * $timeArray[0];
$mins_in_secs = $timeArray[1];
$date = strtotime("midnight", strtotime($dateString));
$timestamp = $date + $hours_in_secs + $mins_in_secs;
//assign timestamp to validation array
$array['display_date'] = $timestamp;
echo $array['display_date']; // Output = 1351094430 (Oct 24 2012 17:00:30)
This really depends on what $dateString contains. Assuming your jQuery widget delivered the time portion as well, your colleague likely wanted to remove the time portion. Compare the following:
echo date(DATE_ATOM, strtotime('2010-10-01 17:32:00'));
// 2010-10-01T17:32:00+02:00
echo date(DATE_ATOM, strtotime("midnight", strtotime('2010-10-01 17:32:00')));
// 2010-10-01T00:00:00+02:00
If your widget doesnt return the time portion, I dont see any reason for setting the date to midnight, because it will be midnight automatically:
echo date(DATE_ATOM, strtotime('2010-10-01'));
// 2010-10-01T00:00:00+02:00
Note that all these are dates in the past and they will result in the given year in the past, not the current year like you say. If they do in your code, the cause must be somewhere else.
Will there be repercussions when you change the code? We cannot know. This is just one line of code and we have no idea of any context. Your unit-tests should tell you when something breaks when you change code.
EDIT after update
The codeblock you show makes no sense whatsoever. Ask the guy who wrote it what it is supposed to do. Not only will it falsely return the current year for past years, but it will also give incorrect results for the minutes, e.g.
24 March, 2010 17:30 will be 2012-03-24T17:00:30+01:00
I assume this was an attempt at turning 24 March, 2010 17:30 into a valid timestamp, which is in a format strtotime does not recognize. But the approach is broken. When you are on PHP5.3 use
$dt = DateTime::createFromFormat('d F, Y H:i', '24 March, 2010 17:30');
echo $dt->format(DATE_ATOM); // 2010-03-24T17:30:00+01:00
If you are not on 5.3 yet, go through https://stackoverflow.com/search?q=createFromFormat+php for alternate solutions. There is a couple in there.
I'm trying to get the number of the month of the year by the number of a week of the year and the year.
So for example week 1 is in january and returns 1, week 6 is in february so I want 2.
I tried to go with date_parse_from_format('W/Y') but had no success (it's giving me errors).
Is there any way to go with date_parse_from_format() or is there another way?
print date("m",strtotime("2011-W6-1"));
(noting that in 2011, January has six weeks so week 6 (by some definitions) is in month 1).
Just wanted to add a note for the first answer, the week number should be 01-09 for Weeks 1 through 9 (it will always give month 1 if you don't add the leading zero)
date("m",strtotime("2011-W06-1"));
Using PHP DateTime objects (which is the preferred way of dealing with dates see links below for more info) you can accomplish it this way:
$dateTime = new \DateTime();
$dateTime->setISODate($year,$week);
$month = $dateTime->format('n');
Note that the following will not work as week "W" is not a supported format:
$month = \DateTime::createFromFormat("W/Y ", "1/2015")->format('n');
The format used by this method is the same supported by the function you where trying to use date_parse_from_format, hence the errors.
Why PHP DateTime Rocks
DateTime class vs. native PHP date-functions
strtotime notes
PHP/Architect's Guide to Date and Time Programming (Chapter 2)
Something like this will do, this is also tested and works:
function getMonthByNumber($number,$year)
{
return date("F",strtotime('+ '.$number.' weeks', mktime(0,0,0,1,1,$year,-1)));
}
echo getMonthByNumber(27,2011);
Hope this helps
I'm preparing a query for mySQL to grab record from the previous week, but I have to treat weeks as Monday - Sunday. I had originally done this:
WHERE YEARWEEK(contactDate) = YEARWEEK(DATE_SUB(CURDATE(),INTERVAL 7 DAY))
to discover that mySQL treats weeks as Sunday - Monday. So instead I'm parsing getting the begin & end dates in php like this:
$i = 0;
while(date('D',mktime(0,0,0,date('m'), date('d')-$i, date('y'))) != "Mon") {
$i++;
}
$start_date = date('Y-n-j', mktime(0,0,0,date('m'), date('d')-($i+7), date('y')));
$end_date = date('Y-n-j', mktime(0,0,0,date('m'), date('d')-($i+1), date('y')));
This works - it gets the current week's date for monday (walking backwards until a monday is hit) then calculates the previous week's dates based on that date.
My question is: Is there a better way to do this? Just seems sloppy, and I expect someone out there can give me a cleaner way to do it - or perhaps not because I need Monday - Sunday weeks.
Edit
Apparently, there is:
$start = date('Y-m-d',strtotime('last monday -7 days'));
$end = date('Y-m-d',strtotime('last monday -1 days'));
That's about a million times more readable. Thank you.
you can use strtotime for this kind of date issues
echo strtotime("last Monday");
(complementing on marvin and Stomped ) Also you can use it this way
echo date('Y-m-d',strtotime('-1 Monday')); //last Monday
echo date('Y-m-d',strtotime('-2 Monday')); //two Mondays ago
echo date('Y-m-d',strtotime('+1 Monday')); //next Monday
strtotime("previous week Monday")
Monday of the previous week.
Marvin's answer is really elegant, although if you really wanted to go for performance you could do it with a little arithmetic. You could derive a formula/method to convert an arbitrary date to "days since some starting point" (probably 01.01.0000) and then the rest of the operations would be easy with that. Getting the day of week from such a number is as simple as subtracting and getting the remainder of a division.
Actually, PHP had a Date class in its PEAR library which did exactly this.
I have to point out for all the readers here there is a big issue with marvin's answer
Here is the catch
The "last Monday" function will return date the "latest" Monday.It will be explained like this , if today is Monday, then it will return the date of "LAST WEEK" Monday, however if today is not Monday, it will return "THIS WEEK" Monday
The Question is request to return "Last Week" Monday. Therefore the result will be incorrect if today is not Monday.
I have solved the issue and wrapped in my Date Time Helper
It will be only one line after you "INCLUDE" the class
$lastWeekMonday = Model_DTHpr::getLastWeekMonday();
Check Here
https://github.com/normandqq/Date-Time-Helper