I have a code that print the date in this format
Sun, 09 Mar 2014 08:31:14 GMT
I want to take the Date and shorten it so I can print
the day
the month
the year
but I want to print each separately and I want to print the month in numbers
Here is the code I tried
$Date = $item->pubDate;
echo '.$Date.';
You should make use of createFromFormat of the DateTimeClass
<?php
$dt='Sun, 09 Mar 2014 08:31:14 GMT';
$date = DateTime::createFromFormat('D, d M Y G:i:s O', $dt);
echo "Day : ".$date->format('d'); // 09
echo "Month : ".$date->format('m'); //03
echo "Year : ".$date->format('Y'); //2014
Demo
Try using date function
echo date('d-m-Y', strtotime($Date));
And if you want to echo one by one
$date_to_time = strtotime($Date);
//Day
echo date('d', $date_to_time);
//Month
echo date('m', $date_to_time);
//Year
echo date('Y', $date_to_time);
Try
$date = '08/02/2014';
list($month, $day, $year) = explode('/', $date);
Related
I am trying to convert a date to month name and year.
$date = '2017-07-00';
$date = date('m/y', strtotime($date));
echo DATE_FORMAT($date, '%M %Y');
I am expecting output like
July, 2017
Here is error i am getting
Warning: date_format() expects parameter 1 to be DateTimeInterface, string given
No Need of DATE_FORMAT() function.
Example-1: If 00 used in day. Then, output will be June, 2017
<?php
$date = '2017-07-00';
echo date('F, Y', strtotime($date)); //June, 2017
?>
Example-2: If 01 or valid day used in day. Then, output will be July, 2017
<?php
$date = '2017-07-01';
echo date('F, Y', strtotime($date)); //July, 2017
?>
You are not using correct parameters, use F for moth and Y for year
Full code:
$date = '2017-07-00';
$date = date('F, Y ', strtotime($date));
echo $date;
Try This.
$date = '2017-07-01';
$date = date('F, Y', strtotime($date));
echo $date;
You can use this
echo date('F,Y',strtotime($date));
I want to change date format like
$date ="21 dec '15";
to
$date ="21-12-2015";
How can it is possible in php.
$date = DateTime::createFromFormat('!m', $result['month']);
echo $date->format('F');
F stands for "month name",
$date = new DateTime(str_replace("'", '20', "21 dec '15"));
var_dump($date->format('d-m-Y'));
Use this
$date = "21 dec '2015"; #your date
$replaced_date = str_replace("'","",$date);
$new = date_create($replaced_date);
$new_date = date_format($new, "d-m-Y"); # restructer the date
echo $new_date ;
Phpfiddle Preview
Suppose I have 2 dates say 29 Aug 2014 and 3 Sep 2014. I need to display all dates between these to dates in the below format.
Aug 2014
29 Fri
30 Sat
31 Sun
Sept 2014
01 Mon
02 Tue
03 Wed
I know how to print all the dates like 29,30,31,1,2,3. But what I am unable to do is to get the month names in between.
Pretty easy question to be honest, pretty basic sollution possible..
$dateRange = new DatePeriod(
new DateTime('2014-07-28'),
new DateInterval('P1D'),
new DateTime('2014-08-04 00:00:01')
);
$month = null;
foreach ($dateRange as $date)
{
$currentMonth = $date->format('m Y');
if ($currentMonth != $month)
{
$month = $date->format('m Y');
echo $date->format('F Y').'<br />';
}
echo $date->format('d D').'<br />';
}
Above sollution results in:
July 2014
28 Mon
29 Tue
30 Wed
31 Thu
August 2014
01 Fri
02 Sat
03 Sun
Do mind it needs PHP >= 5.3 (due to the use of DatePeriod), but the actual logic to solve your question is easy to implement regardless of the used PHP version.
$timeS = strtotime("29 Aug 2014");
$timeE = strtotime("3 Sep 2014");
$monthS = -1;
$time = $timeS;
while ($time < $timeE) {
if ($monthS != date("n", $time)) {
echo date("M Y", $time) . "\n";
$monthS = date("n", $time);
}
echo date("d D", $time) . "\n";
$time = strtotime("+1 day", $time);
}
Edit: After doing it I'm pretty ok with #hindmost comment :)
I think , this is the complete code , as you wanted.
Executed code is here...
http://phpfiddle.org/main/code/3cbe-4855
<?php
$currentMonth = null;
$timeS = strtotime("29 Aug 2013");
$timeE = strtotime("3 Sep 2014");
$time = $timeS;
while ($time < $timeE) {
$month = date("M", $time);
$year = date("Y", $time);
if ($month != $currentMonth)
echo "<br /><h3>".$month."- ".$year."</h3>";
$currentMonth = $month;
echo "<br />".date("d D", $time);
$time = strtotime("+1 day", $time);
}
?>
I have this PHP code:
$end=date('Y-m-d');
I use it to get the current date, and I need the date 5 years in the future, something like:
$end=date('(Y + 5)-m-d');
How can I do this?
Try with:
$end = date('Y-m-d', strtotime('+5 years'));
Modifying dates based on this post
strtotime() is really powerful and allows you to modify/transform dates easily with it’s relative expressions too:
Procedural
$dateString = '2011-05-01 09:22:34';
$t = strtotime($dateString);
$t2 = strtotime('-3 days', $t);
echo date('r', $t2) . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
DateTime
$dateString = '2011-05-01 09:22:34';
$dt = new DateTime($dateString);
$dt->modify('-3 days');
echo $dt->format('r') . PHP_EOL; // returns: Thu, 28 Apr 2011 09:22:34 +0100
The stuff you can throw at strtotime() is quite surprising and very human readable. Have a look at this example looking for Tuesday next week.
Procedural
$t = strtotime("Tuesday next week");
echo date('r', $t) . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
DateTime
$dt = new DateTime("Tuesday next week");
echo $dt->format('r') . PHP_EOL; // returns: Tue, 10 May 2011 00:00:00 +0100
Note that these examples above are being returned relative to the time now.
The full list of time formats that strtotime() and the DateTime constructor takes are listed on the PHP Supported Date and Time Formats page.
Another example, suitable for your case could be: based on this post
<?php
//How to get the day 3 days from now:
$today = date("j");
$thisMonth = date("n");
$thisYear = date("Y");
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+3, $thisYear));
//1 week from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth, $today+7, $thisYear));
//4 months from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+4, $today, $thisYear));
//3 years, 2 months and 35 days from now:
list($today,$thisMonth,$thisYear) = explode(" ", date("j n Y"));
echo date("F j Y", mktime(0,0,0, $thisMonth+2, $today+35, $thisYear+3));
?>
Use this code to add years or months or days or hours or minutes or seconds to a given date
echo date("Y-m-d H:i:s", strtotime("+1 years", strtotime('2014-05-22 10:35:10'))); //2015-05-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 months", strtotime('2014-05-22 10:35:10')));//2014-06-22 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 days", strtotime('2014-05-22 10:35:10')));//2014-05-23 10:35:10
echo date("Y-m-d H:i:s", strtotime("+1 hours", strtotime('2014-05-22 10:35:10')));//2014-05-22 11:35:10
echo date("Y-m-d H:i:s", strtotime("+1 minutes", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:36:10
echo date("Y-m-d H:i:s", strtotime("+1 seconds", strtotime('2014-05-22 10:35:10')));//2014-05-22 10:35:11
You can also subtract replacing + to -
$date = strtotime($row['timestamp']);
$newdate = date('d-m-Y',strtotime("+1 year",$date));
Its very very easy with Carbon.
$date = "2016-02-16"; // Or Your date
$newDate = Carbon::createFromFormat('Y-m-d', $date)->addYear(1);
Using Carbon:
$dt = Carbon::now();
echo $dt->addYears(5);
To add one year to todays date use the following:
$oneYearOn = date('Y-m-d',strtotime(date("Y-m-d", mktime()) . " + 365 day"));
You may use DateInterval for this purpose;
$currentDate = new \DateTime(); //creates today timestamp
$currentDate->add(new \DateInterval('P5Y')); //this means 5 Years
and you can now format it;
$currentDate->format('Y-m-d');
Try below code, i hope it will be helpful for you
<?php
$current_date=strtotime(date('Y-m-d'));
echo $end = date('Y-m-d', strtotime('+5 years',$current_date));
?>
try this ,
$presentyear = '2013-08-16 12:00:00';
$nextyear = date("M d,Y",mktime(0, 0, 0, date("m",strtotime($presentyear )), date("d",strtotime($presentyear )), date("Y",strtotime($presentyear ))+5));
echo $nextyear;
try this:
$yearnow= date("Y");
$yearnext=$yearnow+1;
echo date("Y")."-".$yearnext;
Try this code and add next Days, Months and Years
// current month: Aug 2018
$n = 2;
for ($i = 0; $i <= $n; $i++){
$d = strtotime("$i days");
$x = strtotime("$i month");
$y = strtotime("$i year");
echo "Dates : ".$dates = date('d M Y', "+$d days");
echo "<br>";
echo "Months : ".$months = date('M Y', "+$x months");
echo '<br>';
echo "Years : ".$years = date('Y', "+$y years");
echo '<br>';
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
PHP date() and strtotime() return wrong months on 31st
I have this code and it outputs something strange i think. So, what i am doing wrong here.
<?php
$sP1 = date('m Y');
$sP2 = date('m Y', strtotime('+01 month'));
$sP3 = date('m Y', strtotime('+02 month'));
$sP4 = date('m Y', strtotime('+03 month'));
echo $sP1.'<br>';
echo $sP2.'<br>';
echo $sP3.'<br>';
echo $sP4.'<br>';
?>
and this outputs
05 2012
07 2012
07 2012
08 2012
i think the second one should be
06 2012
Anybody know any solution?
Today is the 31st next month only has 30 days so it would be 7/12 in 1 month from today
assuming that today is May 31 2012
date('m Y') == 05 2012
date('m Y', strtotime('+1 month')) == 07 2012 because june has 30 days
date('m Y', strtotime('+2 month')) == 07 2012
date('m Y', strtotime('+3 month')) == 08 2012
date('m Y', strtotime('+4 month')) == 10 2012
I would take today's date and find the first day of the month then add a month to that if you are doing something that needs to get each month
As others have said, it is because today is the 31st and +1 month equals June-31 which changes to Jul-1. If you include the day in the date string, you can see exactly this.
<?php
$sP1 = date('m-d-Y');
$sP2 = date('m-d-Y', strtotime('+01 month'));
$sP3 = date('m-d-Y', strtotime('+02 month'));
$sP4 = date('m-d-Y', strtotime('+03 month'));
echo $sP1."\n";
echo $sP2."\n";
echo $sP3."\n";
echo $sP4."\n";
/* Outputs:
05-31-2012
07-01-2012
07-31-2012
08-31-2012
*/
?>
strtotime though can take the start date as part of the string so as King suggested, calculate the +N months from the first. So a string like May-1-2012 +01 month such as:
<?php
$sP1 = date('m Y');
$sP2 = date('m Y', strtotime(date('M-1-Y').' +01 month'));
$sP3 = date('m Y', strtotime(date('M-1-Y').' +02 month'));
$sP4 = date('m Y', strtotime(date('M-1-Y').' +03 month'));
echo $sP1."\n";
echo $sP2."\n";
echo $sP3."\n";
echo $sP4."\n";
/* Outputs:
05 2012
06 2012
07 2012
08 2012
*/
?>
http://codepad.org/auYLHvDI
It is working as intended. In a nutshell, it is because what is "one month" from May 31? June 30? August 1?
My suggestion is that if you need sequential months, calculate the offset from the start of the current month, not the current day. Or compose the date that you're looking for manually using the month, day, and year parts broken up.