Hướng dẫn sub year php

I've got a date in this format:

2009-01-01

How do I return the same date but 1 year earlier?

elitalon

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asked Jan 2, 2010 at 1:43

2

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'

answered Jan 2, 2010 at 1:45

0

Use strtotime[] function:

  $time = strtotime["-1 year", time[]];
  $date = date["Y-m-d", $time];

answered Jan 2, 2010 at 1:45

AlexAlex

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1

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'];

answered Sep 10, 2013 at 8:34

darrenwhdarrenwh

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1

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.. :]

answered Oct 26, 2016 at 6:58

saadksaadk

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// 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];

answered Jan 2, 2010 at 1:57

NirmalNirmal

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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.

answered Dec 5, 2018 at 8:09

MelomanMeloman

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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: //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: //www.php.net/manual/en/class.dateinterval.php

answered Jan 13, 2020 at 13:09

Oliver TappinOliver Tappin

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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']];

answered Aug 26, 2017 at 11:37

0

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