I'm trying to convert date (29/04/2017) to day in week but I get wrong value (Thursday, but need to be Saturday). I check the time of the server, all correct.
my code:
$timestamp = strtotime("29/04/2017");
$day = date('l', $timestamp);
echo $day;
what can be the problem?
When you use slashes in your date, strtotime() assumes MM/DD/YYYY. You either need to change the format of your date or use DateTime::createFromFormat() to parse the date
$timestamp = strtotime("04/29/2017");
$day = date('l', $timestamp);
echo $day;
or
$date = DateTime::createFromFormat('d/m/Y, '29/04/2017);
$day = $date->format('l');
echo $day;
Related
I'm trying to extract just the day of the month from the dateTime formatted as follows:
$today = date('Y-m-d\TH:i:sP', strtotime('today'));
But, when I try extracting just the day of the month, I get '31'.
I'm using: $day = date('d', $today);
Which, I'm guessing, is incorrect.
This is because the second parameter to date() needs to be a Unix Timestamp. You're passing it string. As a result you get a date of Dec 31, 1969.
All of that code is unnecessary anyways as all you need is:
$day = day('d');
If you're going to only have access to the date string you must convert it to Unix Timestamp before passing it to date().
To extract days (or other parts of a datetime), I use the format function:
$datetime = new DateTime('2000-01-10', new DateTimeZone('Pacific/Nauru'));
$day = $datetime->format('d');
echo $day;
Use any format form the PHP Manual.
Hope this will solve your problem
$today = date('Y-m-d\TH:i:sP', strtotime('today'));
$day = date('d', strtotime($today)); // here is the difference,
// instead of $today use strtotime($today)
I want to find the next occurrence of date from the current date.
For example, imagine I want to find 20th of the month from current date
If current date is 10th october then return the result 2014-10-20 (Y-m-d)
If current date is 22nd october then return the result 2014-11-20 (Y-m-d)
I created a solution just now using a while loop.
$oldd= "2014-06-20";
$newdate = date("Y-m-d",strtotime($oldd."+1months"));
while(strtotime($newdate) <= strtotime(date("Y-m-d")))
{
$newdate = date("Y-m-d",strtotime($newdate."+1months"));
}
echo $newdate;
manually and pass that to strtotime(). The time information you need to extract from the reference time string. Like this:
$refdate = '2014-02-25 10:30:00';
$timestamp = strtotime($refdate);
echo date('Y-m-d H:i:s',
strtotime("next Thursday " . date('H:i:s', $timestamp), $timestamp)
);
same results could be achieved using string concatenation:
echo date('Y-m-d', strtotime("next Thursday", $timestamp)
. ' ' . date('H:i:s', $timestamp);
another way, you can use methods of DateTime object, PHP has really rich API in dealing with date time.
$current_date = new DateTime('2014-06-20');
if ($current_date->format('d') >= 20) {
// $current_date->modify('last day of this month')->modify("+20 days");
$current_date->modify('first day of next month')->modify("+19 days");
}else{
$current_date->modify('first day of this month')->modify("+19 days");
}
echo $current_date->format("Y-m-d");
http://php.net/manual/en/datetime.modify.php
http://php.net/manual/en/datetime.formats.relative.php
I am generating next date using the following code:
$s1=date('d/M/Y', strtotime('+1 day'));
echo $s1;
for ex: Assume current date is 26/Aug/2014.
so above code generates 27/Aug /2014 and storing in varible $s1.
By using the varible s1 i want to create 28/Aug/2014. how to create?
I dont want to use '+2 day' in STRTOTIME function. I want to generate next day based on variable $s1.
You can do it all with strtotime() but you have to remember that strtotime assumes a USA date format when it see's a / forward slash as a seperator.
So before using $s1 you need to convert the / to a - so it assumes a sensible data format is being used.
$s1=date('d/M/Y', strtotime('+1 day'));
echo $s1.PHP_EOL;
// change date format as strtotime assumes USA dates
$date = strtotime( '+1 day', strtotime( str_replace('/','-',$s1) ) );
echo date('d/M/Y', $date);
When run on 26/Aug/2014 the result would be
27/Aug/2014
28/Aug/2014
The best way (using strtotime):
$tomorrow = strtotime('+1 day');
$twoDaysHence = strtotime('+1 day', $tomorrow);
echo date('d/M/Y', $tomorrow);
echo date('d/M/Y', $twoDaysHence);
In other words, leave your date variables in the form of UNIX timestamps as returned by strtotime until you need to display them. Because you can do calculations directly with them in this format. Once you format that to a date string, you'll have to convert them back into a malleable form first. strtotime doesn't recognise the format d/M/Y automatically, so that makes that all the harder. You should use DateTime in that case:
$tomorrow = date('d/M/Y', strtotime('+1 day'));
$timestamp = DateTime::createFromFormat('d/M/Y', $tomorrow);
$timestamp->modify('+1 day');
echo $timestamp->format('d/M/Y');
You can use something like following:
$newvariable = strtotime ('+2 day' , $s1);
this is a very simple part
$s1=date('d/M/Y', strtotime('+2 day'));
echo $s1;
and if you want then copy the value of $s1 in another variable
function date_addDate($text, $da=0, $ma=0, $ya=0, $ha=0)
{
$h=date('H',strtotime($text));
$d=date('d',strtotime($text));
$m=date('m',strtotime($text));
$y=date('Y',strtotime($text));
$fromTime =date("Y-m-d H:i:s", mktime($h+$ha, 0, 0, $m+$ma, $d+$da, $y+$ya));
return $fromTime;
}
$date = date("Y-m-d H:i:s");
// $da days
// $ma months
// $ya years
// $ha hours
echo date_addDate($date, $da=0, $ma=0, $ya=0, $ha=0);
//out put : current date
echo date_addDate($date, $da=2, $ma=0, $ya=0, $ha=0);
//out put : As you want
Try this
$s1=date('d/M/Y', strtotime('+1 day'));
echo $s1; echo "<br/>";
$date = strtotime(strtotime($s1). ' + 2 days');
$s2 = date('d/M/Y', $date);
echo $s2;
Now it's edited!! Check it!
What about using DateTime ?
$d1 = new DateTime(date('Y-m-d'));
$d1->format('d/m/Y'); // 26/08/2014
$d1->modify('+1 day');
$d1->format('d/m/Y'); // 27/08/2014
Im trying to add a certain amount of days to a timestmp using this in PHP:
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate);
$endDate2 = strtotime('+1 day',$endDate);
echo $endDate2;
but its displaying: 1216526400
any ideas?
Try:
echo date("Y-m-d H:i:s",$endDate2);
Or (for just the date):
echo date("Y-m-d",$endDate2);
You can find documentation about how to format your string here: http://php.net/manual/en/function.date.php
You should be using DateTime for working with dates. It's timezone friendly.
$datetime = new DateTime('2008-06-20');
$datetime->modify('+1 day');
echo $datetime->getTimestamp();
strtotime() converts the date into a unix timestamp which is the number of seconds since January 1st 1970. If you want a date output you have to run the finished timestamp through date() first.
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate.' +1 day');
echo date("Y-m-d", $endDate);
strtotime creates a Unix timestamp so if you want to be presented with a formatted date, you need to pass the timestamp as an argument to the date function as follows:
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate);
$endDate2 = strtotime('+1 day',$endDate);
echo date('Y-m-d', $endDate2);
Additionally, there are a wide variety of parameters you can use in the date function if you want to display additional information.
e.g.: echo date('Y-m-d H:i:s', $endDate2); or echo date('Y-m-d h:i:s a', $endDate2);, etc.
Sooooo close, just take your timestamp and convert it back into date format using date("desired format",$endDate2);
DateTime is a very nice way to deal with dates. You can try like this:
$capturedDate = '2008-06-20';
$date = DateTime::createFromFormat('Y-m-d', $capturedDate)->modify('+1 day');
echo $date->getTimestamp();
I've got a date in this format:
2009-01-01
How do I return the same date but 1 year earlier?
You can use strtotime:
$date = strtotime('2010-01-01 -1 year');
The strtotime function returns a unix timestamp, to get a formatted string you can use date:
echo date('Y-m-d', $date); // echoes '2009-01-01'
Use strtotime() function:
$time = strtotime("-1 year", time());
$date = date("Y-m-d", $time);
Using the DateTime object...
$time = new DateTime('2099-01-01');
$newtime = $time->modify('-1 year')->format('Y-m-d');
Or using now for today
$time = new DateTime('now');
$newtime = $time->modify('-1 year')->format('Y-m-d');
an easiest way which i used and worked well
date('Y-m-d', strtotime('-1 year'));
this worked perfect.. hope this will help someone else too.. :)
On my website, to check if registering people is 18 years old, I simply used the following :
$legalAge = date('Y-m-d', strtotime('-18 year'));
After, only compare the the two dates.
Hope it could help someone.
// set your date here
$mydate = "2009-01-01";
/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastyear = strtotime("-1 year", strtotime($mydate));
// format and display the computed date
echo date("Y-m-d", $lastyear);
Although there are many acceptable answers in response to this question, I don't see any examples of the sub method using the \Datetime object: https://www.php.net/manual/en/datetime.sub.php
So, for reference, you can also use a \DateInterval to modify a \Datetime object:
$date = new \DateTime('2009-01-01');
$date->sub(new \DateInterval('P1Y'));
echo $date->format('Y-m-d');
Which returns:
2008-01-01
For more information about \DateInterval, refer to the documentation: https://www.php.net/manual/en/class.dateinterval.php
You can use the following function to subtract 1 or any years from a date.
function yearstodate($years) {
$now = date("Y-m-d");
$now = explode('-', $now);
$year = $now[0];
$month = $now[1];
$day = $now[2];
$converted_year = $year - $years;
echo $now = $converted_year."-".$month."-".$day;
}
$number_to_subtract = "1";
echo yearstodate($number_to_subtract);
And looking at above examples you can also use the following
$user_age_min = "-"."1";
echo date('Y-m-d', strtotime($user_age_min.'year'));