Create Datetime object from string

Question

I have a situation where I need to create a Datetime object from a string. The issue comes when I don't have consistency in the pattern of the string representing the date. These are examples of my data:

As you can see, the last one is invalid no matter what, because there is no 17 month, but the first 3 are valid depending on the month position(and of course, the year)

My question: is there any way(pretty sure through regex) to make a function with the date string as parameter that return the datetime object. If the string is not valid, return: 1/1/1970...

You have to think for a solution for how your program can differentiate the different formats. — moffeltje
– moffeltje, Commented Jun 4, 2015 at 14:45
For example: count the number of characters before the first '/'. If it is 4 you know its the year. — moffeltje
– moffeltje, Commented Jun 4, 2015 at 14:46

dognose · Accepted Answer · 2015-06-04 14:59:07Z

The issue comes when I don't have consistency in the pattern of the string representing the date

for this purpose, php offers strtotime(): http://php.net/manual/en/function.strtotime.php

Example usage:

   $str1 = "2015-06-04 16:00";
   $str2 = "06/04/2015 4 pm";
   $str3 = "04.06.2015 16:00";

   $actualDate = date("Y-m-d H:i:s", strtotime($str1));
   echo $actualDate."<br />";
   $actualDate = date("Y-m-d H:i:s", strtotime($str2));
   echo $actualDate."<br />";
   $actualDate = date("Y-m-d H:i:s", strtotime($str3));
   echo $actualDate."<br />";

   //all will produce "2015-06-04 16:00:00"

as a bonus, strtotime also supports "insane" expressions like

$actualDate = date("Y-m-d H:i:s", strtotime("06/04/2015 + 1 day - 8 hours")); 
echo $actualDate."<br />";
// "2015-06-04 16:00:00"

and many more such as "monday this week", "tuesday next week", "first monday of january 2038" and the like.

konradwww · Accepted Answer · 2015-06-04 14:46:36Z

1

You can try to create DateTime Object with your string values. It will throw exception if the date format is not valid, then you can just catch it and return your 1/1/1971

try {
    $dateTime = new DateTime('17/17/2015');
    return $dateTime;
} catch (Exception $e) {
   return '1/1/1971';
}

answered Jun 4, 2015 at 14:46

konradwww

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Comments

cjhill · Accepted Answer · 2015-06-04 14:39:47Z

0

You can use DateTime.

$myDate = '17/17/2015';

$date = DateTime::createFromFormat('d/m/Y', $myDate);

if (DateTime::getLastErrors()['warning_count'] >= 1) {
    // Invalid
} else {
    // Valid date
}

answered Jun 4, 2015 at 14:39

cjhill

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3 Comments

Oriam Over a year ago

The problem is I can't asure this format: 'd/m/Y' in createFromFormat('d/m/Y', $myDate) Take a look to the data examples

cjhill Over a year ago

You could use strtotime, then?

cjhill Over a year ago

If you allow completely custom dates you will always encounter problems. E.g., 1/2/15 might be 1st February or the 2nd January.

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