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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?

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4
  • 1
    because its 0000-00-00 in your db Commented Jan 24, 2012 at 9:39
  • its not coming from db, its from my html page to php Commented Jan 24, 2012 at 9:41
  • The strtotime input is wrong. Check the $date1. Commented Jan 24, 2012 at 9:41
  • 2
    @AssamGuy - The format dd/mm/yyyy should be dd-mm-yyyy so you will need to replace your / characters. Commented Jan 24, 2012 at 9:42

8 Answers 8

Reset to default
87

Replace / with -:

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

Simple and effective @Cyclone!
I had this probem when coverting two UK dates to MySQL format. Your help is very much appreciated. Works like a dream!
I do the same but problem is still can you help @Cyclonecode
@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()?
36

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.

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1 Comment

Thus, putting a condition likeif (strtotime($date1)) before the conversion seems like a good practice.
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

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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.

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1 Comment

I've used this with PHPExcel and it works perfect. Thank you!
0

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;
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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.

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-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');
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1 Comment

I don't see a difference between those two
-2

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

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

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?

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