PHP subtract 1 month from date formatted with date ('m-Y') - php

Question

I'm trying to subtract 1 month from a date.
$today = date('m-Y');
This gives: 08-2016
How can I subtract a month to get 07-2016?

Answer 1

<?php
echo $newdate = date("m-Y", strtotime("-1 months"));
output
07-2016

Answer 2

Warning! The above-mentioned examples won't work if call them at the end of a month.
<?php
$now = mktime(0, 0, 0, 10, 31, 2017);
echo date("m-Y", $now)."\n";
echo date("m-Y", strtotime("-1 months", $now))."\n";
will output:
10-2017
10-2017
The following example will produce the same result:
$date = new DateTime('2017-10-31 00:00:00');
echo $date->format('m-Y')."\n";
$date->modify('-1 month');
echo $date->format('m-Y')."\n";
Plenty of ways how to solve the issue can be found in another thread: PHP DateTime::modify adding and subtracting months

Answer 3

Try this,
$today = date('m-Y');
$newdate = date('m-Y', strtotime('-1 months', strtotime($today)));
echo $newdate;

Answer 4

Depending on your PHP version you can use DateTime object (introduced in PHP 5.2 if I remember correctly):
<?php
$today = new DateTime(); // This will create a DateTime object with the current date
$today->modify('-1 month');
You can pass another date to the constructor, it does not have to be the current date. More information: http://php.net/manual/en/datetime.modify.php

Answer 5

I used this to prevent the "last days of month"-error. I just use a second strtotime() to set the date to the first day of the month:
<?php
echo $newdate = date("m-Y", strtotime("-1 months", strtotime(date("Y-m")."-01")));

Answer 6

if(date("d") > 28){
$date = date("Y-m", strtotime("-".$loop." months -2 Day"));
} else {
$date = date("Y-m", strtotime("-".$loop." months"));
}

Answer 7

$lastMonth = date('Y-m', strtotime('-1 MONTH'));

Answer 8

First change the date format m-Y to Y-m
$date = $_POST('date'); // Post month
or
$date = date('m-Y'); // currrent month
$date_txt = date_create_from_format('m-Y', $date);
$change_format = date_format($date_txt, 'Y-m');
This code minus 1 month to the given date
$final_date = new DateTime($change_format);
$final_date->modify('-1 month');
$output = $final_date->format('m-Y');

Answer 9

Try this,
$effectiveDate = date('2018-01'); <br>
echo 'Date'.$effectiveDate;<br>
$effectiveDate = date('m-y', strtotime($effectiveDate.'+-1 months'));<br>
echo 'Date'.$effectiveDate;

Answer 10

$currentMonth = date('m', time());
$currentDay = date('d',time());
$currentYear = date('Y',time());
$lastMonth = $currentMonth -1;
$one_month_ago=mkdate(0,0,0,$one_month_ago,$currentDay,$currentYear);
This could be rewritten more elegantly, but it works for me

Answer 11

I realize this is an old post, but I've been solving the same issue, and here is what I came up with to account for all the variability. This function is just trying to get relative dates, so same day of prior month, or last day of month if you are on the last day, regardless of exactly how many days a month has. So goal is given '2010-03-31' and subtract a month, we should output '2010-02-28'.
private function subtractRelativeMonth(DateTime $incomingDate): DateTime
{
$year = $incomingDate->format('Y');
$month = $incomingDate->format('m');
$day = $incomingDate->format('d');
$daysInOldMonth = cal_days_in_month(CAL_GREGORIAN, $month, $year);
if ($month == 1) { //It's January, so we have to go back to December of prior year
$month = 12;
$year--;
} else {
$month--;
}
$daysInNewMonth = cal_days_in_month(CAL_GREGORIAN, $month, $year);
if ($day > $daysInNewMonth && $month == 2) { //New month is Feb
$day = $daysInNewMonth;
}
if ($day > 29 && $daysInOldMonth > $daysInNewMonth) {
$day = $daysInNewMonth;
}
$adjustedDate = new \DateTime($year . '-' . $month . '-' . $day);
return $adjustedDate;
}