I have a date/time string like this: 180510_112440 in this format ddmmyy_hhmmss
I need a snippet for having a string formatted like this way: 2010-05-18 11:24:40
Thanks for help.
another possible answer is the common use of strptime to parse your date and the mktime function:
<?php
$orig_date = "180510_112440";
// Parse our date in order to retrieve in an array date's day, month, etc.
$parsed_date = strptime($orig_date, "%d%m%y_%H%M%S");
// Make a unix timestamp of this parsed date:
$nice_date = mktime($parsed_date['tm_hour'],
$parsed_date['tm_min'],
$parsed_date['tm_sec'],
$parsed_date['tm_mon'] + 1,
$parsed_date['tm_mday'],
$parsed_date['tm_year'] + 1900);
// Verify the conversion:
echo $orig_date . "\n";
echo date('d/m/y H:i:s', $nice_date);
$inDate = '180510_112440';
$date = strtotime('20'.substr($inDate,4,2).'-'.
substr($inDate,2,2).'-'.
substr($inDate,0,2).' '.
substr($inDate,7,2).':'.
substr($inDate,9,2).':'.
substr($inDate,11,2));
echo date('d-M-Y H:i:s',$date);
Assumes date will always be in exactly the same format, and always 21st century
list($d,$m,$y,$h,$i,$s)=sscanf("180510_112440","%2c%2c%2c_%2c%2c%2c");
echo "20$y-$m-$d $h:$i:$s";
Related
I have a string: 30/06/18 (30th June 2018)
I am converting to a date:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result: 18-06-2018
Now I want to add 20 days to the date:
$expiryDate = date("d-m-Y", strtotime("+20 days", $calcFieldDate));
echo $expiryDate;
Expected Result: 08-07-2018
Actual Result: 31-01-1970
I am obviously creating a date format which is then subsequently being treated as a string...
Every time I try a conversion, I just hit another road block - is there anyway to create a date that is then treated like a date?
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result:30-06-2018
$expiryDate = date("d-m-Y", strtotime("+20 days", strtotime($calcFieldDate)));
echo $expiryDate;
Result:20-07-2018
Strtotime() The second parameter is the timestamp
You actually don't need to revert using strtotime and date functions, you can actually use DateTime to simply add dates into it:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18');
echo $calcFieldDate->format('d-m-Y'); // get inputted date
$expiryDate = clone $calcFieldDate; // clone the original date object
$expiryDate->modify('+20 days'); // adjust the cloned date
echo $expiryDate->format('d-m-Y'); // show the adjusted date
This will sort your problem.
$str="30/06/18 (30th June 2018)";
$arr_temp=explode(" ",$str);
$str_date=str_replace("/","-",$arr_temp[0]);
$dt = DateTime::createFromFormat('d-m-y',$str_date);
$date=$dt->format('d-m-Y');
$new_date=date('d-m-Y',strtotime("+20 days",strtotime($date)));
echo $new_date;
I have a string that contains date and time. Format of my string is yyyymmddtime.
For example. 20171125123000209. this is my complete string in which first comes the month, then month and then day after that time. How can i retrieve date from it by converting to readable date format. I tried with php's date function. But the output was not as expected. Please help.
Try this
<?php
$str_date= "20171125123000209";
$exiting_date_format='Ymd';
//first 8 characters from given date string in second parameter below
$date = DateTime::createFromFormat($exiting_date_format, substr($str_date,0,8));
echo $date->format('Y-m-d');//specify desired date format
?>
Output :
2017-11-25
DateTime::createFromFormat - Parses a time string according to a specified format
You can use date() function of php
<?php
$full_date= "20171125123000209";
echo date('dS F h:i:s A', $full_date);
//Output = 31st May 12:30:09 AM
?>
Or,
To get date from timestamp like now,
$timestamp= time(); //Or your timestamp here
$date = date('d-m-Y', $timestamp); //Inside first parameter, give your date format
echo $date; //17-12-2017
Or,
To get anything from string you can also use substr() function of php.
<?php
$full_date= "20171125123000209";
$year = substr($full_date, 0, 4);
$month = substr($full_date, 4, 2);
$date = substr($full_date, 6, 2);
echo 'Year = '.$year.' ';
echo 'Month = '.$month.' ';
echo 'Date = '. $date.' ';
?>
Output:
Year = 2017 Month = 11 Date = 25
Test in jdoodle
About substr() function in php
It's not necessary to manipulate the string at all. PHP's DateTime class supports parsing a string containing miliseconds natively, using the u format modifier:
$str = '20171125123000209';
$date = DateTime::createFromFormat('YmdHisu', $str);
Using your new $date object, you can convert to whatever format you're looking for, e.g.
echo $date->format("F j Y, g:i a");
// November 25 2017, 12:30 pm
See https://eval.in/920636
$your_strtotime_val = ""; //20171125123000209
$convert_to_date = date("m-d-Y h:i:s",$your_strtotime); // [m-d-Y h:i:s] this depending on how you convert date in first time so be careful
I have a variable is which the value coming is Date along with time in php. How do I convert it into a variable to get only the year? I do not need automatic updation but the format change is needed. Normal answers are giving it about date but my variable is containing time as well.
The format coming by now is 2017-12-11 4:06:37 and i need only 2017
Use like this:
<?php echo date('Y',strtotime('now'));?>
You can you simple DateTime function and date_formate() function for displaying separate year, month and date.
For that you have to first convert in Object of your current Date time string by using :
$date = new \DateTime('2017-12-11 4:06:37');
And then you can use date format function by using below code:
echo date_format($date, "Y"); //for Display Year
echo date_format($date, "m"); //for Display Month
echo date_format($date, "d"); //for Display Date
You can code like this (working perfectly):
$format = 'Y-m-d H:i:s';
$date = DateTime::createFromFormat($format, '2009-02-15 15:16:17');
echo "Format: $format; " . $date->format('Y') . "\n";
As mentioned by Himanshu Upadhyay, this is correct and the easiest way.
<?php
echo date('Y',strtotime('now'));
?>
But i would recommend you to read this here. You should really do actually!
By using DateTime class
$date = new \DateTime('2017-12-11 4:06:37');
echo $date->format('Y');
Hi I know this question is a bit trivial. I googled it but could not find the exact problem anywhere.
I have a string which contains a application filling date how can I convert this string into date
$appln_filling_date = '20020315';
The type of appln_filling_date is string I want to convert its type to date and want the data remain the same. The field in the database has a type date.
EDIT
This is what I am trying to do
$appln_filling_date = strtotime($appln_data['bibliographic-data']['application-reference']['document-id']['1']['date']['$']);
$appln_filling_date = date('Y-m-d', $appln_filling_date);
If your date is in some standard date format use this example:
$d = new DateTime('20020315');
echo $d->format('Y-m-d');
# or [if you do not have DateTime, which is in php >= 5.2.0]
$t = strtotime('20020315');
echo date('Y-m-d', $t);
If your date is not in some standard format use this example:
$d = DateTime::createFromFormat('d . Y - m', '15 . 2002 - 03');
echo $d->format('Y-m-d');
$date = DateTime::createFromFormat('Ymd', $appln_filling_date);
This worked for me like a charm
$appln_filling_date = $appln_data['bibliographic-data']['application-reference']['document-id']['1']['date']['$'];
$appln_filling_date = date_create(date('Y-m-d', $appln_filling_date));
Thanks for the down voting
I'm having date 20/12/2001 in this formate . i need to convert in following format 2001/12/20 using php .
$var = explode('/',$date);
$var = array_reverse($var);
$final = implode('/',$var);
Your safest bet
<?php
$input = '20/12/2001';
list($day, $month, $year) = explode('/',$input);
$output= "$year/$month/$day";
echo $output."\n";
Add validation as needed/desired. You input date isn't a known valid date format, so strToTime won't work.
Alternately, you could use mktime to create a date once you had the day, month, and year, and then use date to format it.
If you're getting the date string from somewhere else (as opposed to generating it yourself) and need to reformat it:
$date = '20/12/2001';
preg_replace('!(\d+)/(\d+)/(\d+)!', '$3/$2/$1', $date);
If you need the date for other purposes and are running PHP >= 5.3.0:
$when = DateTime::createFromFormat('d/m/Y', $date);
$when->format('Y/m/d');
// $when can be used for all sorts of things
You will need to manually parse it.
Split/explode text on "/".
Check you have three elements.
Do other basic checks that you have day in [0], month in [1] and year in [2] (that mostly means checking they're numbers and int he correct range)
Put them together again.
$today = date("Y/m/d");
I believe that should work... Someone correct me if I am wrong.
You can use sscanf in order to parse and reorder the parts of the date:
$theDate = '20/12/2001';
$newDate = join(sscanf($theDate, '%3$2s/%2$2s/%1$4s'), '/');
assert($newDate == '2001/12/20');
Or, if you are using PHP 5.3, you can use the DateTime object to do the converting:
$theDate = '20/12/2001';
$date = DateTime::createFromFormat('d/m/Y', $theDate);
$newDate = $date->format('Y/m/d');
assert($newDate == '2001/12/20');
$date = Date::CreateFromFormat('20/12/2001', 'd/m/Y');
$newdate = $date->format('Y/m/d');