PHP date_format(): How to format date from string

2019-05-05 03:29发布

I have a bit of PHP code:

$exd = date_create('01 Dec, 2015');
$exd = date_format($exd, 'Y-m-d');
echo $exd;

Which is used for formatting the date. The expected output would be 2015-12-01 but it returns 2016-12-01. What am i missing?

4条回答
虎瘦雄心在
2楼-- · 2019-05-05 03:33

Use createFromFormat method first, provide the input format:

$exd = DateTime::createFromFormat('d M, Y', '01 Dec, 2015');
// arguments (<format of the input>, <the input itself>)
$exd = date_format($exd, 'Y-m-d'); // then choose whatever format you like
echo $exd;
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SAY GOODBYE
3楼-- · 2019-05-05 03:45

i got the solution of your bug that is date_format(datae_variable,date_format);

<?php
     $exd = date_create('01 Dec, 2015');
     $exd1 = date_format($exd,"Y-m-d");//here you make mistake
     echo $exd1;
?>
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Animai°情兽
4楼-- · 2019-05-05 03:52

It can be a date function call simply. Use stringtotime for exact/precise date/time value

date("Y-m-d",strtotime("01 Dec 2015"))

When you run this code the out put will show

2015-12-01

this is because of the comma in the string which terminates the date string in the compiler. If you specify exactly the timezone (like $timezone = 'America/New_York) . parameter you can show precise time as well.

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你好瞎i
5楼-- · 2019-05-05 03:54

The date_create() function accepts only the parameter link, This function is also and alias function of DateTime::__construct()

check the function date_create_from_format() its also a alias function of DateTime::createFromFormat(). Refer link

$exd = date_create_from_format('j M, Y', '01 Dec, 2015');
//$exd = date_create('01 Dec, 2015');
$exd = date_format($exd, 'Y-m-d');
echo $exd;
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