CODEWITHSUNDEEP
php
JezB
JezB

Reputation: 528

PHP Date Formats

This should be simple, but I'm having trouble... so turn to StackOverflow...

I'm in the UK and am getting a date from the jQuery DatePicker in a dd/mm/yy format. I want to store this date as a serial (yyyymmdd) so run a Date_To_Serial function that just does:

return date("Ymd", strtotime($strDate_In));

where $strDate_In is the date string in dd/mm/yy format.

So, passing in 01/12/2013, I expect 20131201 to be returned.

But, when 01/12/2013 is passed in, it returns 20130112 - PHP obviously assumes the date format is mm/dd/yyyy.

How can I fix this please?

Upvotes: 0

Views: 130

Answers (4)

Michael
Michael

Reputation: 12802

If the separator is a / then strtotime assumes a US-format. To have it recognise a UK format the separator must be - or .:

echo date('Ymd', strtotime('01/12/2014')); // 20140112
echo date('Ymd', strtotime('01-12-2014')); // 20141201

So for this to work in your example you would do:

return date("Ymd", strtotime(str_replace('/', '-', $strDate_In)));

Upvotes: 1

Maximilian Riegler
Maximilian Riegler

Reputation: 23506

Check out DateTime::createFromFormat for correct handling of non-standard date formats.

return DateTime::createFromFormat("d/m/Y", $strDate_In)->format("Ymd");

Upvotes: 0

Madara's Ghost
Madara's Ghost

Reputation: 174937

Use a DateTime object's createFromFormat method. It allows you to specify the format of the input.

Afterwards, use the format method to export the date in the desired format.

Upvotes: 0

Kasper Pedersen
Kasper Pedersen

Reputation: 516

If you know the format, try using the static createFromFormat method on the DateTime class:

$date = DateTime::createFromFormat('d/m/Y', '01/12/2013');
return $date->format('Ymd');

Upvotes: 5

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