Why PHP is returning wrong days between two dates?

Question

There are a ton of such questions on Stackoverflow and before you mark this as duplicate, let me tell you that none of those work in my case.

Date Difference in php on days?

Calculate the difference between date/times in PHP

Dates difference with php

Finding the number of days between two dates

How to calculate the difference between two dates using PHP?

Above are the questions I have checked. The last question has an answer by Jurka which also unfortunately doesn't works. So here is my code:

$date1 = new DateTime(date('d-m-y'));
$date2 = new DateTime($etaDate);
$interval = $date1->diff($date2);
echo "difference " . $interval->days . " days ";

The $etaDate in above case is 03-02-16 and the current date is 08-02-16 so it should return 5 days but it returns a very high figure - 1826 days. I have tried different methods which also returns the same day figure.

A date value is stored as string in database so on retrieving I am again putting it back to date format using this code:

$newdate = DateTime::createFromFormat('d-m-y', $EntryDate);
$newdate->add(new DateInterval('P15D'));
$etaDate = $newdate->format('d-m-y');

$EntryDate is a variable that holds the date from the database, $etaDate converts it into date format.

So what format is `03-02-16` and `08-02-16` - Is that 16th February 2003 and 16th February 2008? If you feed ambiguous values into PHP's date functions, then you may not get the expected outputs — Mark Baker, Feb 08 '16 at 10:56
`$etaDate` considered `2003-02-16`. How are you getting that date? — Sougata Bose, Feb 08 '16 at 10:57
How is PHP expected to know the date format if a human being can't say for certain what date it is — Mark Baker, Feb 08 '16 at 10:58
Then you need to tell PHP that is the format that you're using: `$date1 = new DateTime(); $date2 = new DateTime::createFromFormat('d-m-y', $etaDate);` ` — Mark Baker, Feb 08 '16 at 11:10
@MarkBaker That's what I did, you can see that in the last part of the question. — gegobyte, Feb 08 '16 at 11:12
I have answered solution here: https://stackoverflow.com/a/63060175/219112 — Sadee, Jul 23 '20 at 17:47

score 1 · Accepted Answer · answered Feb 08 '16 at 11:01

1

Just change the format of year from y to Y and it will work.

See the example : https://eval.in/514811

answered Feb 08 '16 at 11:01

Mr. Engineer

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Using 'd-m-y', I am inserting into database so I have to use 'y' only. 16 cannot be compared to 2016. – gegobyte Feb 08 '16 at 11:04
You can always change your date format when you want to insert into database. – Mr. Engineer Feb 08 '16 at 11:05
Yeah that is true but what is the problem with 'd-m-y' format because it is also valid. – gegobyte Feb 08 '16 at 11:06
Problem is in `03-02-16`. How can PHP identify `year` in this format? – Mr. Engineer Feb 08 '16 at 11:08
In the last part of question, I have mentioned how I arrived at that date, using this: `$etaDate = $newdate->format('d-m-y');` So PHP knows I am using this format. – gegobyte Feb 08 '16 at 11:13
It is really risky task to use two digit format for year. i suggest you to use four digit. How will you know that `16` is `2016` or `1916`? – Mr. Engineer Feb 08 '16 at 11:31
Yeah never thought about that, changed `y` to `Y` and now it works. – gegobyte Feb 08 '16 at 11:40

score 0 · Answer 2 · answered Feb 08 '16 at 11:06

0

Try This Code //YYYY-MM-DD format

$etaDate="2016-02-03";
$date1 = new DateTime(date('Y-m-d'));
$date2 = new DateTime($etaDate);
$interval = $date1->diff($date2);
echo "difference " . $interval->days . " days ";

answered Feb 08 '16 at 11:06

Vadivel S

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Why PHP is returning wrong days between two dates?

2 Answers2