PHP date showing '1970-01-01 ' after conversion

Question

I have a form in which date format is dd/mm/yyyy . For searching database , I hanverted the date format to yyyy-mm-dd . But when I echo it, it showing 1970-01-01 . The PHP code is below:

$date1 = $_REQUEST['date'];     
echo date('Y-m-d', strtotime($date1));

Why is it happening? How can I format it to yyyy-mm-dd?

@AssamGuy - The format `dd/mm/yyyy` should be `dd-mm-yyyy` so you will need to replace your `/` characters. — Cyclonecode, Jan 24 '12 at 09:42

Cyclonecode · Accepted Answer · 2016-05-05T02:12:36.030

87

Replace / with -:

$date1 = strtr($_REQUEST['date'], '/', '-');
echo date('Y-m-d', strtotime($date1));

edited May 05 '16 at 02:12

answered Jan 24 '12 at 09:40

Cyclonecode

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Simple and effective @Cyclone! – Neil May 04 '16 at 14:44
1

I had this probem when coverting two UK dates to MySQL format. Your help is very much appreciated. Works like a dream! – Nick Green Mar 10 '17 at 06:17
1

I do the same but problem is still can you help @Cyclonecode – always-a-learner May 30 '17 at 11:24
1

@ankitsuthar - Very hard to give a solution without seeing the date you're trying to format. Try removing any whitespaces from the string before using `strtotime()`? – Cyclonecode Jun 09 '17 at 20:44

score 36 · Answer 2 · answered Jan 24 '12 at 09:43

January 1, 1970 is the so called Unix epoch. It's the date where they started counting the Unix time. If you get this date as a return value, it usually means that the conversion of your date to the Unix timestamp returned a (near-) zero result. So the date conversion doesn't succeed. Most likely because it receives a wrong input.

In other words, your strtotime($date1) returns 0, meaning that $date1 is passed in an unsupported format for the strtotime function.

Thus, putting a condition like`if (strtotime($date1))` before the conversion seems like a good practice. — joeljpa, Mar 02 '23 at 11:17

score 2 · Answer 3 · answered Oct 24 '18 at 12:12

2

$inputDate = '07/05/-0001';
$dateStrVal = strtotime($inputDate);
if(empty($dateStrVal))
{
  echo 'Given date is wrong'; 
}
else{
 echo 'Date is correct';
}

O/P : Given date is wrong

answered Oct 24 '18 at 12:12

Kaushal Roy

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1
6

score 1 · Answer 4 · edited Oct 16 '17 at 20:13

1

$date1 = $_REQUEST['date'];

if($date1) {
    $date1 = date( 'Y-m-d', strtotime($date1));
} else {
    $date1 = '';
}

This will display properly when there is a valid date() in $date and display nothing if not.
Solved the issue for me.

edited Oct 16 '17 at 20:13

Axel

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answered Oct 16 '17 at 16:46

Tye Lucas

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I've used this with PHPExcel and it works perfect. Thank you! – Bogdan Stoica Feb 03 '22 at 19:03

score 0 · Answer 5 · answered Dec 12 '18 at 06:12

Another workaround:

Convert datepicker dd/mm/yyyy to yyyy-mm-dd

$startDate = trim($_POST['startDate']);
$startDateArray = explode('/',$startDate);
$mysqlStartDate = $startDateArray[2]."-".$startDateArray[1]."-".$startDateArray[0];
$startDate = $mysqlStartDate;

score 0 · Answer 6 · answered Feb 11 '20 at 21:04

0

The issue is when your data is set to 000-00-00 or empty you must double-check and give the correct information and this issue will go away. I hope this helps.

answered Feb 11 '20 at 21:04

jerryurenaa

3,863
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score -1 · Answer 7 · answered Apr 29 '17 at 06:34

-1

Use below code for php 5.3+:

$date = new DateTime('1900-02-15');
echo $date->format('Y-m-d');

Use below code for php 5.2:

$date = new DateTime('1900-02-15');
echo $date->format('Y-m-d');

answered Apr 29 '17 at 06:34

Dayz

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7

I don't see a difference between those two – Eyeslandic Jun 17 '17 at 10:02

score -2 · Answer 8 · answered Aug 19 '15 at 06:29

-2

finally i have found a one line code to solve this problem

date('d/m/Y', strtotime(str_replace('.', '-', $row['DMT_DATE_DOCUMENT'])));

answered Aug 19 '15 at 06:29

Bakhtawar GIll

407
5
16

2

What kind of answer is this, you picked code from everyone else and put it on one line? Have you ever heard about code readability? – Cyclonecode Feb 09 '16 at 13:36

PHP date showing '1970-01-01 ' after conversion

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