convert month from name to number

Question

Is there an easy way to change $month = "July"; so that $nmonth = 7 (07 would be fine too). I could do a case statement, but surely there is already a function to convert? EDIT: I wish I could accept multiple answers, cause two of you basically gave me what I needed by your powers combined.

$nmonth = date('m',strtotime($month));

That will give the numerical value for $month. Thanks!

score 93 · Answer 1 · answered Jul 19 '10 at 17:48

93

Try this:

<?php
  $date = date_parse('July');
  var_dump($date['month']);
?>

answered Jul 19 '10 at 17:48

Matthew

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this should be #1.... simple one-liner – James Walker Mar 24 '23 at 01:47

score 78 · Accepted Answer · answered Jul 19 '10 at 17:46

78

Yes,

$date = 'July 25 2010';
echo date('d/m/Y', strtotime($date));

The m formats the month to its numerical representation there.

answered Jul 19 '10 at 17:46

Sarfraz

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score 31 · Answer 3 · answered Mar 30 '12 at 11:10

An interesting look here, the code given by kelly works well,

$nmonth = date("m", strtotime($month));

but for the month of february, it won't work as expected when the current day is 30 or 31 on leap year and 29,30,31 on non-leap year.It will return 3 as month number. Ex:

$nmonth = date("m", strtotime("february"));

The solution is, add the year with the month like this:

$nmonth = date("m", strtotime("february-2012"));

I got this from this comment in php manual.

amazing....! Never thought about leap trouble years... – gumuruh Feb 10 '22 at 07:09 — gumuruh, Feb 10 '22 at 07:09

score 13 · Answer 4 · answered Aug 20 '13 at 11:04

13

$string = "July";
echo $month_number = date("n",strtotime($string));

returns '7' [month number]

Use date("m",strtotime($string)); for the output "08"

For more formats reffer this..
http://php.net/manual/en/function.date.php

answered Aug 20 '13 at 11:04

V A S

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score 5 · Answer 5 · edited Sep 13 '12 at 18:50

5

you can also use this one:

$month = $monthname = date("M", strtotime($month));

edited Sep 13 '12 at 18:50

eckes

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answered Sep 13 '12 at 18:34

lokeshpahal

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Kelly Copley · Answer 6 · 2010-07-19T18:16:42.673

4

$nmonth = date("m", strtotime($month));

edited Jul 19 '10 at 18:16

answered Jul 19 '10 at 17:57

Kelly Copley

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I tested this solution and in some cases it returns the wrong number, e.g. '5' for 'April'. Anybody got an idea why? I'll have a closer look at it later. – Patrick Mar 31 '12 at 01:57
@longeasy Have you passed the month along with the year, have a look at my answer below. – vkGunasekaran Mar 31 '12 at 06:21

Vipin Choubey · Answer 7 · 2019-11-18T07:14:47.780

4

If you want number of month from string name then

$month = 'August';
$year = 2019;
echo date('m',strtotime($month.' '.$year));

Gives 08

Or If you want the Full name of the month then

echo date('F')

OR if you want the half name of the month then

echo date('M')

edited Nov 18 '19 at 07:14

answered Aug 19 '19 at 06:15

Vipin Choubey

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If you want the number of the month without leading zeroes, use 'n' (so `echo date('n', strtotime($month.' '.$year));`) – qozle Oct 13 '22 at 19:18

score 4 · Answer 8 · edited Mar 31 '12 at 14:17

4

$monthname = date("F", strtotime($month));

F means full month name

edited Mar 31 '12 at 14:17

Juan Mellado

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answered Mar 31 '12 at 10:42

Ranjith Siji

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This returns the full month name, whereas the OP is looking for the month number. Replacing the "F" with "m" (for two digit month number i.e. '05' for May) or "n" (for single digit month number i.e. "5" for May) yields the result that OP is looking for. – qozle Oct 13 '22 at 19:13

score 3 · Answer 9 · answered Dec 20 '12 at 11:12

3

<?php
$monthNum = 5;
$monthName = date("F", mktime(0, 0, 0, $monthNum, 10));
echo $monthName; //output: May
?>

answered Dec 20 '12 at 11:12

Nikos Tsirakis

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OP is looking for the number of the month, not the month name. They're trying to translate from month name to month number. "F" in `date()` returns the month name...use 'm' or 'n' for the month number with or without leading zeroes (respectively) – qozle Oct 13 '22 at 19:17

score 2 · Answer 10 · edited Nov 16 '19 at 08:47

Above answer is good. Here is another way to do:-

function getMonthNumber($monthStr) {
//e.g, $month='Jan' or 'January' or 'JAN' or 'JANUARY' or 'january' or 'jan'
$m = ucfirst(strtolower(trim($monthStr)));
switch ($m) {
    case "January":        
    case "Jan":
        $m = "01";
        break;
    case "February":
    case "Feb":
        $m = "02";
        break;
    case "March":
    case "Mar":
        $m = "03";
        break;
    case "April":
    case "Apr":
        $m = "04";
        break;
    case "May":
        $m = "05";
        break;
    case "June":
    case "Jun":
        $m = "06";
        break;
    case "July":        
    case "Jul":
        $m = "07";
        break;
    case "August":
    case "Aug":
        $m = "08";
        break;
    case "September":
    case "Sep":
        $m = "09";
        break;
    case "October":
    case "Oct":
        $m = "10";
        break;
    case "November":
    case "Nov":
        $m = "11";
        break;
    case "December":
    case "Dec":
        $m = "12";
        break;
    default:
        $m = false;
        break;
}
return $m;
}

February, not "Febuary" – Evgeny Fedin Mar 05 '18 at 07:54 — Evgeny Fedin, Mar 05 '18 at 07:54

Ben Guthrie · Answer 11 · 2010-07-19T18:31:21.670

2

It may be easiest to create a fake date so you can use the date function.

Excellent reference here: http://php.net/manual/en/function.date.php

Example:

<?
$month = 7;

$tempDate = mktime(0, 0, 0, $month, 1, 1900); 

echo date("m",$tempDate);


?>

edited Jul 19 '10 at 18:31

answered Jul 19 '10 at 17:50

Ben Guthrie

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score 1 · Answer 12 · answered May 13 '13 at 10:21

1

Use

date("F", mktime(0, 0, 0, ($month)));

where, $month value will be 1 -> 12

answered May 13 '13 at 10:21

PHP Ferrari

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OP wanted the number of the month, not the name. Use "m" or "n" in place of "F" (in `date("F", ...`) to get the month number with or without leader zeroes (respectively). – qozle Oct 13 '22 at 19:20

score 1 · Answer 13 · edited Jun 08 '17 at 09:30

1

$date = 'Dec 25 2099';
echo date('d/m/Y', strtotime($date));

This returns 01/01/1970, that means php doesn't support all dates, it returns correct formatted date till Jan 19 2038 but Jan 20 2038 returns 01/01/1970

edited Jun 08 '17 at 09:30

Muhammad Hassaan

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answered Nov 05 '13 at 07:06

Haneefa

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score 1 · Answer 14 · edited Dec 06 '12 at 11:09

1

Maybe use a combination with strtotime() and date()?

edited Dec 06 '12 at 11:09

Sergey Berezovskiy

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answered Jul 19 '10 at 17:47

robertbasic

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A complete working example here as a solution would be helpful. – qozle Oct 13 '22 at 19:20

score 1 · Answer 15 · answered Jul 09 '17 at 07:52

1

$dt = '2017-Jan-10';
         OR
$dt = '2017-January-10';

echo date('Y-m-d', strtotime($dt));

echo date('Y/m/d', strtotime($dt));

answered Jul 09 '17 at 07:52

Mohammad Adil

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score 0 · Answer 16 · answered Jun 12 '13 at 23:59

0

With PHP 5.4, you can turn Matthew's answer into a one-liner:

$date = sprintf('%d-%d-01', $year, date_parse('may')['month']);

answered Jun 12 '13 at 23:59

Mike Purcell

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score -2 · Answer 17 · answered Jun 11 '15 at 10:11

I know this might seem a simple solution, but why not just use something like this

<select name="month">
  <option value="01">January</option>
  <option value="02">February</option>
  <option selected value="03">March</option>
</select>

The user sees February, but 02 is posted to the database

convert month from name to number

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