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 ga开发者_StackOverflowve me what I needed by your powers combined.
$nmonth = date('m',strtotime($month));
That will give the numerical value for $month
.
Thanks!
Try this:
<?php
$date = date_parse('July');
var_dump($date['month']);
?>
Yes,
$date = 'July 25 2010';
echo date('d/m/Y', strtotime($date));
The m
formats the month to its numerical representation there.
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.
$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
you can also use this one:
$month = $monthname = date("M", strtotime($month));
$nmonth = date("m", strtotime($month));
$monthname = date("F", strtotime($month));
F
means full month name
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')
<?php
$monthNum = 5;
$monthName = date("F", mktime(0, 0, 0, $monthNum, 10));
echo $monthName; //output: May
?>
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);
?>
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 $m;
}
Maybe use a combination with strtotime()
and date()
?
Use
date("F", mktime(0, 0, 0, ($month)));
where, $month value will be 1 -> 12
$date = 'Dec 25 2099';
echo date('d/m/Y', strtotime($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
$dt = '2017-Jan-10';
OR
$dt = '2017-January-10';
echo date('Y-m-d', strtotime($dt));
echo date('Y/m/d', strtotime($dt));
With PHP 5.4, you can turn Matthew's answer into a one-liner:
$date = sprintf('%d-%d-01', $year, date_parse('may')['month']);
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
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