php detect and convert dates from a string - php

Question

I was wondering if I could somehow to detect a date in a string and convert it into a standard date format.
Let's consider the input strings below:
Company registered on 16 March 2003
or
Activity between 10 May 2006 an 10 July 2008 - no changes.
Now I would like a PHP function to apply over the strings and get the dates as YYYY-mm-dd
Example:
$date = DateExtract($sting1); // output: 2003-03-16
$date = DateExtract($sting2); // output: ['2006-05-10','2008-07-10']

Answer 1

For finding first two digit number Date Regexp would be - (?<![0-9])[0-9]{2}(?![0-9]) This can also be apply to four digit for Year also and for Month you can use hard-coded string search code.
$string = "Activity between 10 May 2006 an 10 July 2008 - no changes.";
preg_match_all('/(\d{1,2}) (\w+) (\d{4})/', $string, $matches);
print_r($matches);
Output :
Array
(
[0] => Array
(
[0] => 10 May 2006
[1] => 10 July 2008
)
[1] => Array
(
[0] => 10
[1] => 10
)
[2] => Array
(
[0] => May
[1] => July
)
[3] => Array
(
[0] => 2006
[1] => 2008
)
)
Updates
For find Complete Date in string you can use this -
It works for short code for Month like Jan and complete name like January also.
Code:
$string = "Activity between 10 May 2006 an 10 July 2008 - no changes.";
preg_match_all('/(\b\d{1,2}\D{0,3})?\b(?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sep(?:tember)?|Oct(?:ober)?|(Nov|Dec)(?:ember)?)\D?(\d{1,2}\D?)?\D?((19[7-9]\d|20\d{2})|\d{2})/', $string, $complete);
print_r($complete);
Result:
Array
(
[0] => Array
(
[0] => 10 May 2006
[1] => 10 July 2008
)
[1] => Array
(
[0] => 10
[1] => 10
)
[2] => Array
(
[0] =>
[1] =>
)
[3] => Array
(
[0] => 20
[1] => 20
)
[4] => Array
(
[0] => 06
[1] => 08
)
[5] => Array
(
[0] =>
[1] =>
)
)
So you can fetch complete date form here and convert it into standard date format.
Demo Here

Answer 2

Tricky. One approach might be to reason that dates always appear after certain grammatical words, as shown in your examples, e.g. "between", "on" etc. Using such words as a beginning anchor, we would then match until we find what we can reasonably assume to be the end of the date string. Here's what I hacked together:
//some strings
$strs = [
"Company was in business between 14 March 2008 and 21 November 2012 inclusive",
"I was born on 29 May 1980, 17:37 - it was a Thursday",
"The big bang did not occur at 2pm, 14 Jun 1971, that's for sure."
];
//container to store possible date matches from strings
$possible_dates = array();
//prep months - long and short forms, to be used in matching
$date_prefix_words = array('between', 'on', 'at', 'during', 'and');
$months = array('January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December');
$months_short = array_map(function($month) { return substr($month, 0, 3); }, $months);
//iterate over and search strings - convert times like 2pm to 14:00, as, if they appear before the date, e.g. string 3, it doesn't get parsed
foreach($strs as $str) {
$str = preg_replace_callback('/\b\d{1,2}[ap]m\b/', function($time) { return date('H:i', strtotime($time[0])); }, $str);
preg_match_all('/(?<=\b'.implode('\b( |:)|\b', $date_prefix_words).'\b( |:))(\d|am|pm| |,|\'|:|'.implode('|', $months).'|'.implode('|', $months_short).')+/i', $str, $matches);
if (count($matches)) $possible_dates = array_merge($possible_dates, $matches[0]);
}
//output before and after results
foreach($possible_dates as &$pd) {
$pd = preg_replace('/, ?$/', '', $pd);
echo '<p>Before: '.$pd.'<br />After: '.date('Y-m-d', strtotime($pd)).'</p>';
}
Clearly I'm making certain assumptions about your date formats, and you may need to tweak the REGEX, but it sort of works.

Answer 3

First of all You have to extract all part of date from the string separately.
RegEx Demo
First approach:
<?php
function standard_date_format($str) {
preg_match_all('/(\d{1,2}) (\w+) (\d{4})/', $str, $matches);
foreach ( $matches[1] as $day ) { $days [] = $day; }
foreach ( $matches[2] as $month ) { $months[] = $month; }
foreach ( $matches[3] as $year ) { $years [] = $year; }
$all_months = array('January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December');
for ($i = sizeof ($days) - 1; $i >= 0; $i--) {
$month = array_search ($months[$i], $all_months) + 1;
$month = strlen ($month) < 2 ? '0'.$month : $month;
$results[] = $years[$i] . '-' . $month . '-' . $days[$i];
}
return $results;
}
$str1 = "Company registered on 16 March 2003";
$str2 = "Activity between 10 May 2006 an 10 July 2008 - no changes.";
print_r(standard_date_format($str1)); // output: 2003-03-16
print_r(standard_date_format($str2)); // output: ['2006-05-10','2008-07-10']
PHP Demo
Second approach:
<?php
function standard_date_format($str) {
preg_match_all('/(\d{1,2}) (\w+) (\d{4})/', $str, $matches);
$dates = array_map("strtotime", $matches[0]);
$result = array_map(function($v) {return date("Y-m-d", $v); }, $dates);
return $result;
}
$str1 = "Company registered on 16 March 2003";
$str2 = "Activity between 10 May 2006 an 10 July 2008 - no changes.";
print_r(standard_date_format($str1)); // output: 2003-03-16
print_r(standard_date_format($str2)); // output: ['2006-05-10','2008-07-10']
PHP Demo