PHP - checking if two dates match but ignoring the year - php

I have an array which will output a date. This date is outputted in the mm/dd/yyyy format. I have no control over how this outputted so I cant change this.
Array
(
[date] => 04/06/1989
)
I want to use php to check if this date matches the current date (today), but ignoring the year. So in the above example I just want to check if today is the 6th April. I am just struggling to find anything which documents how to ignore the years.

if( substr( $date, 0, 5 ) == date( 'm/d' ) ) { ...
Works only if it's certain that the month and date are both two characters long.

Came in a little late, but here’s one that doesn’t care what format the other date is in (e.g. “Sep 26, 1989”). It could come in handy should the format change.
if (date('m/d') === date('m/d', strtotime($date))) {
echo 'same as today';
} else {
echo 'not same as today';
}

this will retrieve the date in the same format:
$today = date('m/d');

Use this:
$my_date = YOUR_ARRAY[date];
$my_date_string = explode('/', $my_date);
$curr_date = date('m,d,o');
$curr_date_string = explode(',', $date);
if (($my_date_string[0] == $curr_date_string[0]) && ($my_date_string[1] == $curr_date_string[1]))
{
DO IT
}
This way, you convert the dates into strings (day, month, year) which are saved in an array. Then you can easily compare the first two elements of each array which contains the day and month.

You can use for compare duple conversion if you have a date.
$currentDate = strtotime(date('m/d',time())); --> returns current date without care for year.
//$someDateTime - variable pointing to some date some years ago, like birthday.
$someDateTimeUNIX = strtotime($someDateTime) --> converts to unix time format.
now we convert this timeunix to a date with only showing the day and month:
$dateConversionWithoutYear = date('m/d',$someDateTimeUNIX );
$dateWithoutRegardForYear = strtotime($dateConversionWithoutYear); -->voila!, we can now compare with current year values.
for example: $dateWithoutRegardForYear == $currentDate , direct comparison

You can convert the other date into its timestamp equivalent, and then use date() formatting to compare. Might be a better way to do this, but this will work as long as the original date is formatted sanely.
$today = date('m/Y', time());
$other_date = date('m/Y', strtotime('04/06/1989'));
if($today == $other_date) {
//date matched
}

hi you can just compare the dates like this
if(date('m/d',strtotime($array['date']])) == date('m/d',strtotime(date('Y-m-d H:i:s',time()))) )

Related

PHP: How to know if a date is in the current month?

I need to know if a date is in the current month.
Examples:
If the date is 2018-06-30 and current month is June (06), then true.
If the date is 2018-07-30 and current month is June (06), then false.
I have a list of dates with more than 1000 dates and I want to show or colorize only the dates that belongs to a current month.
You can do it all on one line. Basically convert the date in question to a PHP time, and get the month.
date('m',strtotime('2018-06-30' )) == date('m');
Using the date() function, if you pass in only the format, it'll assume the current date/time. You can pass in a second optional variable of a time() object to use in lieu of the current date/time.
I hope this helps -
$date = "2018-07-31";
if(date("m", strtotime($date)) == date("m"))
{
//if they are the same it will come here
}
else
{
// they aren't the same
}
As an alternative you could use a DateTime and for the format use for example the n to get the numeric representation of a month without leading zeros and use Y to get the full numeric representation of a year in 4 digits.
$d = DateTime::createFromFormat('Y-m-d', '2018-06-30');
$today = new DateTime();
if($d->format('n') === $today->format('n') && $d->format('Y') === $today->format('Y')) {
echo "Months match and year match";
}
Test
PHP doesn't implement a date type. If you are starting with a date/time and you know that your you are only dealing with a single timezone, AND you mean you want the current month in the curent year
$testdate=strtotime('2018-06-31 12:00'); // this will be converted to 2018-07-01
if (date('Ym')==date('Ym', $testdate)) {
// current month
} else {
// not current month
}

Check if a date is equal to other date using Laravel 5.3

I am trying to check if one date is equal than the other date, but I can't get the match because the date format coming from the form turns into a different order once it gets through the "parse" code.
I need to format this date to find the match, here is a sample code to show how I am trying:
...
// $ago will give me this date: 2016-12-09 00:00:00
$ago = Carbon\Carbon::today()->addDays(2); // Todays date + 2 days
//$request->datex has the date coming from a form with this format, '12-06-2016'.
// Once a parse $request->datex here, the date gets out of order:
$my_date = Carbon\Carbon::parse($request->datex);
// it shows the date like this, 2016-09-12 00:00:00 , I need it to be on this format: 2016-12-09 00:00:00
// then I could do this:
if ( $ago$ == $my_date ) {
dd($my_date.' is equal to: '.$ago );
}else{
dd(' Not equal!');
}
...
Thanks for looking!
Change this line
$my_date = Carbon\Carbon::parse($request->datex);
with this:
$my_date = Carbon::createFromFormat('m-d-Y', $request->datex)
I've assumed that your format '12-06-2016' means DAY-MONTH-YEAR
UPDATE
Tested my solution on my machine and it works, date is recognized properly:
When
$request->datex = '12-06-2016'
then
$my_date = \Carbon\Carbon::createFromFormat('m-d-Y', $datex);
gives me date like that: public 'date' => string '2016-12-06 18:52:09.000000' (length=26)
Date has been parsed properly. The thing that I've assumed just now. These dates won't be same cause of hours, minutes, seconds and milliseconds. To fix that just we have to compare dates that way:
if ( $ago->format('Y-m-d') == $my_date->format('Y-m-d') )
//do something awesome with our equal dates
PHP expects DD-MM-YYYY or MM/DD/YYYY formats.
If you always have a MM-DD-YYYY format, you could do this before parsing:
$request->datex = str_replace('-', '/', $request->datex);

How to check whether the date coming in $_POST array is not greater than today's date in PHP?

The date to be checked is as follows :
$submission_date = 12-25-2014; //The date in mm-dd-yyyy format that is to be tested against today's date
Now I want to echo the error message since the date contained in a variable $submission_date is a future date.
How should I do this efficiently and effectively using PHP?
Thanks in advance.
Many ways to do this (use DateTime::createFromFormat() to control exact format of input dates, for example) but perhaps the simplest that suits the example is:
$isFuture = (strtotime($submission_date) > strtotime($_POST['current_date']))
Note that OP changed the question. If desired date to test against is not in $_POST array, just replace strtotime($_POST['current_date']) with time() to use current system time.
To compare against current date, disregarding time of day, use:
$today = new DateTime(date("Y-m-d"));
// $today = new DateTime("today"); // better solution courtesy of Glavić
// see http://php.net/manual/en/datetime.formats.relative.php for more info
$today_timestamp = $today->getTimestamp();
If posted format is in m-d-Y, then you cannot convert it to unix timestamp directly with strtotime() function, because it will return false.
If you need to use strtotime() then change the input format to m/d/Y by simple str_replace().
On the other hand, you could use DateTime class, where you can directly compare objects:
$submission_date = DateTime::createFromFormat('!m-d-Y', $submission_date);
$today_date = new DateTime('today');
if ($submission_date > $today_date) {
echo "submission_date is in the future\n";
}
demo
With PHP DateTime you can check whether the input date is future or old w.r.to the todate.
$submission_date = DateTime::createFromFormat('m-d-Y', $submission_date);
$submission_date = $submission_date->format('Y-m-d');
$current_date = new DateTime('today');
$current_date = $current_date->format('Y-m-d');
if ($submission_date > $current_date)
{
echo "Future date";
}
else
{
echo "Old date";
}

Adding leading zeroes to a string date in PHP

I have a string "date" which can be DD.MM.YYYY or D.M.YYYY (with or without leading zeros), it depends what a user types.
Then I use it in a condition to send another email when the day is today.
if($_POST["date"]== date("d.m.Y")){
$headers.="Bcc: another#mail.cz\r\n";
}
The problem is that the mail is send when the date format DD.MM.YYYY (with leading zeros) only.
My proposed solution
As I'm not very good in PHP, I only know the solution theoretically but not how to write the code - I would spend a week trying to figure it out on my own.
What's in my mind is dividing the date into three parts (day, month, year), then checking the first two parts if there's just one digit and adding leading zeros if it's the case. I don't know how to implement that to the condition above, though. I have read a few topics about how to do this, but they were a bit more different than my case is.
You should equalize to same format d.m.Y and you can do this with strtotime and date function:
$post_date = date("d.m.Y", strtotime($_POST["date"]));
if($post_date == date("d.m.Y")){
$headers.="Bcc: another#mail.cz\r\n";
}
I changed date to $post_date for more clear. I'll try to explain difference with outputs
echo $_POST["date"]; // lets say: 8.7.2013
echo date("d.m.Y"); // 09.09.2013 > it's current day
strtotime($_POST["date"]); // 1373230800 > it's given date with unix time
$post_date = date("d.m.Y", strtotime($_POST["date"])); // 08.07.2013 > it's given date as right format
If you use date function without param, it returns as current date.
Otherwise if you use with param like date('d.m.Y', strtotime('given_date'));, it returns as given date.
$post_date = date("d.m.Y", strtotime($_POST["date"]));
At first, we converted your date string to unix with strtotime then equalized and converted format that you used in if clause.
first set date format with leading Zero
$postdate = strtotime('DD.MM.YY', $_POST['date']);
and also matching date will be in same format
$matching_date = date('DD.MM.YY', strtotime('whatever the date'));
then
if ( $postdate === $matching_date )
{
// send mail
}
Why don't you just check the length of the _POST (it can be either 8 or 10)
if (strlen($_POST["date"]) == 10) {
$headers.="Bcc: another#mail.cz\r\n";
}

How to determine if a date is more than three months past current date

I am getting a date back from a mysql query in the format YYYY-MM-DD.
I need to determine if that is more than three months in the past from the current month.
I currently have this code:
$passwordResetDate = $row['passwordReset'];
$today = date('Y-m-d');
$splitCurrentDate = explode('-',$today);
$currentMonth = $splitCurrentDate[1];
$splitResetDate = explode('-', $passwordResetDate);
$resetMonth = $splitResetDate[1];
$diferenceInMonths = $splitCurrentDate[1] - $splitResetDate[1];
if ($diferenceInMonths > 3) {
$log->lwrite('Need to reset password');
}
The problem with this is that, if the current month is in January, for instance, giving a month value of 01, and $resetMonth is November, giving a month value of 11, then $differenceInMonths will be -10, which won't pass the if() statement.
How do I fix this to allow for months in the previous year(s)?
Or is there a better way to do this entire routine?
Use strtotime(), like so:
$today = time(); //todays date
$twoMonthsLater = strtotime("+3 months", $today); //3 months later
Now, you can easily compare them and determine.
I’d use PHP’s built-in DateTime and DateInterval classes for this.
<?php
// create a DateTime representation of your start date
// where $date is date in database
$resetDate = new DateTime($date);
// create a DateIntveral representation of 3 months
$passwordExpiry = new DateInterval('3M');
// add DateInterval to DateTime
$resetDate->add($passwordExpiry);
// compare $resetDate to today’s date
$difference = $resetDate->diff(new DateTime());
if ($difference->m > 3) {
// date is more than three months apart
}
I would do the date comparison in your SQL expression.
Otherwise, PHP has a host of functions that allow easy manipulation of date strings:
PHP: Date/Time Functions - Manual

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