I have a string that is a 10 digit date/time representation in the format of yymmddhhmm, e.g:
1304282240
is the 28th of April, 2013 22:40 (28/04/13 in the DD/MM/YY format)
I need to convert this into a valid PHP timestamp but I'm getting stuck using the Date and strtotime functions, and not sure if that is the best approach in any case.
Update: I've added these lines to my code:
$date = DateTime::createFromFormat('ymdHi', '1304282240');
echo $date->format('m/d/Y h:i:s A');
appears to work
I believe you want something like this:
$date = DateTime::createFromFormat('ymdHi', '1304282240');
Related
I'm working with an XML document that is returning variables and for some reason in a xml return the timestamp is formatted like this... 20180606T110000 ... why anyone would format it like that makes no sense to me; however, its what I have to work with. ITs formatted YYYYMMDD , the T is the split between date and time, HHMMSS. ITs set up in a 24 Hour clock that I also need to convert to 12 hr clock with am/pm
I need that formatted like 06/06/2018 11:00:00 AM.
Is there a way to do that via a date format (I know how to use date() but I don't know how to bring in that timestamp the way its formatted) or even separating it out into
$year = xxxx
$month = xx
$day = $xx
$Hour=xx
etc. etc. etc.
if need be.
I've briefly looked at php's date create from format ( date_create_from_format('j-M-Y', '15-Feb-2009') ) but dont fully understand how that works.
I've also thought about a split. I've also looked at chunk_split and wordwrap but its not even amounts of characters so that would be complex to create.
Any ideas?
The format you're working with is "XMLRPC (Compact)" format. This is fully supported by PHP (you can see a list of supported formats here). To get what you want, just use a combination of strtotime() and date().
$timestring = "20180606T110000";
$timestamp = strtotime($timestring);
echo date("m/d/Y h:i:s A", $timestamp);
You can use PHP DateTime to parse a datetime String with any format. Please view the Parameters format in the following link to understand how the "Ymd\THis" part works: http://php.net/manual/en/datetime.createfromformat.php
<?php
$time = "20180606T110000";
$date = DateTime::createFromFormat("Ymd\THis", $time);
// 06/06/2018 11:00:00 AM.
echo $date->format("d/m/Y h:i:s A");
I have a xml file, containing several dates, in this format: 2016-07-23T07:00:00.000Z. I'm using a php function to convert this in to a format for publishing on a website. This should actually result in something like Saturday, 24th of July (24th, not 23rd, because of the time offset. My function somehow ignores the T07:00:00.000Z part and thus returns Friday, 23rd of July. Can anybody help me out with the proper way to convert this date?
Thanks, Peter
The string in question
2016-07-23T07:00:00.000Z
is a W3C datetime format (W3C DTF) (Complete date plus hours, minutes, seconds and a decimal fraction of a second) which can be properly parsed incl. the fractions of a second with the date_create_from_format](http://php.net/date_create_from_format) function:
$originalDate = "2016-07-23T07:00:00.000Z";
date_create_from_format('Y-m-d\TH:i:s.uO', $originalDate);
It does create a new DateTime which then can be formatted with the for PHP standard codes, e.g.
date_create_from_format('Y-m-d\TH:i:s.uO', $originalDate)
->format('Y-m-d H:i:s'); # 2016-07-23 07:00:00
As that W3C format carries the timezone already and it is UTC, and you wrote you want a different one, you need to specify it:
date_create_from_format('Y-m-d\TH:i:s.uO', $originalDate)
->setTimezone(new DateTimeZone('Asia/Tokyo'))
->format('Y-m-d H:i:s');
The reason why this is not visible (and controlable with the code given) in the previous answer is because date formats according to the default set timezone in PHP where as each DateTime has it's individual timezone.
An equivalent with correct parsing (incl. decimal fraction of a second) with the other answers then is:
$dateTime = date_create_from_format('Y-m-d\TH:i:s.uO', $originalDate);
date('Y-m-d H:i:s', $dateTime->getTimestamp());
Hope this explains it a bit better in case you need the complete date value and / or more control on the timezone.
For the format, see as well: In what format is this date string?
$oldDateTime= "2016-07-23T07:00:00.000Z"; // Your datetime as string, add as variable or whatever.
$newDateTime= date("Y-m-d H:i:s", strtotime($originalDate));
I need to convert a string into date format, but it's returning a weird error. The string is this:
21 nov 2012
I used:
$time = strtotime('d M Y', $string);
PHP returned the error:
Notice: A non well formed numeric value encountered in index.php on line 11
What am I missing here?
You're calling the function completely wrong. Just pass it
$time = strtotime('21 nov 2012')
The 2nd argument is for passing in a timestamp that the new time is relative to. It defaults to time().
Edit: That will return a unix timestamp. If you want to then format it, pass your new timestamp to the date function.
To convert a date string to a different format:
<?php echo date('d M Y', strtotime($string));?>
strtotime parses a string returns the UNIX timestamp represented. date converts a UNIX timestamp (or the current system time, if no timestamp is provided) into the specified format. So, to reformat a date string you need to pass it through strtotime and then pass the returned UNIX timestamp as the second argument for the date function. The first argument to date is a template for the format you want.
Click here for more details about date format options.
You are using the wrong function, strtotime only return the amount of seconds since epoch, it does not format the date.
Try doing:
$time = date('d M Y', strtotime($string));
For more complex string, use:
$datetime = DateTime::createFromFormat("d M Y H:i:s", $your_string_here);
$timestamp = $datetime->getTimestamp();
I need to convert this date:
10.04.2011 19:00
To a date variable that I can use in PHP.
Can someone help me with that? I tried this way:
$dateConverted = date("d.m.Y H:i",strtotime ($date));
But it returns 01.01.1970 00:00
DateTime::createFromFormat() to the rescue!
It looks like your format is d.m.Y H:i.
So, this should work for you:
$dt = DateTime::createFromFormat('d.m.Y H:i', '10.04.2011 19:00');
echo $dt->format('Y-m-d H:i:s');
You should also take a look at the formats that strtotime and DateTime operate on. In particular, the reason that date didn't parse in strtotime is that it only expects dots as delimiters between Y, M and D if the year is only two digits. That's an odd one, don't look at me, it's not my fault.
$data['user']['time'] = '2011-03-07 00:33:45';
how can we add 1 year to this date ?
something like $newdata = $data['user']['time'] + 1 year ?
or
$newdata = 2012-03-07 00:33:45
Thanks
Adam Ramadhan
strtotime() is the function you're looking for:
$data['user']['seal_data'] = date('Y-m-d H:i:s', strtotime('+1 year', strtotime($data['user']['time'])));
First, you have to convert the MySQL datetime to something that PHP can understand. There are two ways of doing this...
Use UNIX_TIMESTAMP() in your query to tell MySQL to return a UNIX timestamp of the datetime column.
SELECT whatever, UNIX_TIMESTAMP(myTime) AS 'myUnixTime' FROM myTable;
Use DateTime::createFromFormat to convert your string time to something PHP can understand.
$date = DateTime::createFromFormat('Y-m-d H:i:s', $data['user']['time']);
Once that is done, you can work with the time... Depending on the method you used above, you can use one of the following.
If you have a unix timestamp, you can use the following to add a year:
$inAYear = strtotime('+1 year', $data['user']['unixTime']);
If you have a DateTime object, you can use the following:
$inAYear = $date->add(new DateInterval('P1Y'));
Now, to display your date in a format that is respectable, you must tell PHP to return a string in the proper format.
If you have a unix timestamp, you can use the following:
$strTime = date('Y-m-d H:i:s', $inAYear);
If you have a DateTime object, you can use the following:
$strTime = $inAYear->format('Y-m-d H:i:s');
Alternatively, if you don't want to deal with all of that, you can simply add one year when you query.
SELECT whatever, DATE_ADD(myTime, INTERVAL 1 YEAR) AS 'inAYear' FROM myTable;
Current (2017) Practice is to use DateTime
This question is top on a google search for "php datetime add one year", but severely outdated. While most of the previous answers will work fine for most cases, the established standard is to use DateTime objects for this instead, primarily due strtotime requiring careful manipulation of timezones and DST.
TL;DR
Convert to DateTime: $date = new DateTime('2011-03-07 00:33:45', [user TZ]);
Use DateTime::modify: $date->modify('+1 year');
Format to needs.
Change the timezone with DateTime::setTimezone from the list of supported timezones: $date->setTimezone(new DateTimeZone('Pacific/Chatham'));
Convert to string with DateTime::format: echo $date->format('Y-m-d H:i:s');
Following this pattern for manipulating dates and times will handle the worst oddities of timezone/DST/leap-time for you.
Just remember two final notes:
Life is easier with your system timezone set at UTC.
NEVER modify the system timezone outside of configuration files.
I've seen too much code that relies on date_default_timezone_set. If you're doing this, stop. Save the timezone in a variable, and pass it around your application instead, please.
More Reading
How to calculate the difference between two dates using PHP?
Convert date format yyyy-mm-dd => dd-mm-yyyy
PHP - strtotime, specify timezone
I think you could use strtotime() to do this pretty easily. Something like:
$newdata = date('c', strtotime($data['user']['time'] . ' +1 year'));
Though the 'c' format string isn't the same as your input format. You could consult date()'s docs for how to construct the correct one.
'Y-m-d H:i:s' — as Tim Cooper suggests — looks correct.
This should do the trick (not tested).
$data = "2011-03-07 00:33:45";
echo 'Original date +1 year: ' . date('Y-m-d H:i:s', strtotime(date("Y-m-d H:i:s", strtotime($data)) . " +1 year"));
First-of-all if your date format is separated by a slash (/), like '2019/12/31' then you should convert it in dash (-) format, like '2019-12-31', to do so use str_replace() function.
$string = str_replace('/', '-', '2019/12/31'); //output: 2019-12-31
To add time/day/month/year do not use strtotime() function, because it can't add a time which is beyond year 2038.
So here I would prefer to use DateTime() function.
$string = '2000-01-01';
$date = new DateTime($string);
$date->add(new DateInterval('P60Y5M2DT6H3M25S')); //60 Years 5 Months 2 Days 6 Hours 3 Minutes 25 Seconds
echo $date->format('Y-m-d H:i:s'); //output: 2060-06-03 06:03:25