Adding leading zeroes to a string date in PHP - 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";
}

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
}

php strtotime() values not working as expected

this code keeps telling me that $lasUpdate is always greater than $yesterday no matter the change i make to $yesterday result is (12/31/14 is greater than 01/19/15 no update needed). i feel like i'm missing something simple thank you in advance it is greatly appreciated.
$result['MAX(Date)']='12/31/14';
$lastUpdate = date('m/d/y', strtotime($result['MAX(Date)']));
$yesterday = date('m/d/y', strtotime('-1 day'));
if($lastUpdate<$yesterday){echo $lastUpdate.'is less '.$yesterday.'<br>'.'update needed';}
if($lastUpdate>=$yesterday){echo $lastUpdate.'is greater than '.$yesterday.'<br>'.'no update needed';
You have fallen victim to PHP type juggling with strings. A date function has a return value of a string. You cannot compare dates in their string format since PHP will juggle strings into integers in the context of a comparison. The only exception is if the string is a valid number. In essence, you are doing:
if ('12/31/14' < '01/19/15') { ... }
if ('12/31/14' >= '01/19/15') { ... }
Which PHP type juggles to:
if (12 < 1) { ... }
if (12 >= 1) { ... }
And returns false on the first instance, and true on the second instance.
Your solution is to not wrap date around the strtotime functions, and just use the returned timestamps from the strtotime functions themselves to compare UNIX timestamps directly:
$lastUpdate = strtotime($result['MAX(Date)']);
$yesterday = strtotime('-1 day');
You will however want to use date when you do the echo back to the user so they have a meaningful date string to work with.
Try something like this:
$lastUpdate = strtotime($result['MAX(Date)']);
$yesterday = strtotime('-1 day');
if ($lastUpdate < $yesterday) { /* do Something */ }
12/31/14 is greater than 01/19/15
Because 1 is greater than 0. If you want to compare dates that way you will need to store them in a different format (from most to least significant digit), for example Ymd.
Or store the timestamps you are making in the different variables and compare them.

What's the issue in a code written for comparing the date with today's date?

I'm comparing a date with current date(i.e. today's date). It is expected that the error should come only when the date to be compared is greater than today's date. It should not come for date which is less than or equal to today's date.
I've written following code for it.
$submission_date = $_POST['submission_date']; //The date in mm-dd-yyyy format that is to be tested against today's date. The value in $submission date is 12-25-2014
//This is a future date. Today's date is 12-10-2014 in dd-mm-yyyy format
$current_date = date('m-d-Y');
if (strtotime($submission_date) > strtotime($current_date))
{
echo "Future date not accepted";
}
With the above code I'm not getting errors for future dates, sometimes I'm getting error for previous dates as well.
How to optimize and make this code correct and standard?
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
If you need to extract some information from DateTime objects, use format() method on them, which accepts same format as date() function:
echo $today_date->format('m/d/Y'); # 12/11/2014
echo $today_date->format('m-d-Y'); # 12-11-2014
echo $today_date->format('Y-m-d'); # 2014-12-11
echo $today_date->format('Y-Y-Y'); # 2014-2014-2014
demo
I think you need to compare date in 'Y-m-d' or 'd-m-Y' format. I think it is not possible to compare date in 'm-d-Y' format
As we have made changes in your code and we test it works from my side so can you try below.
$submission_date = $_POST['submission_date'];
$current_date = date('d-m-y H:i:s');
if (strtotime($submission_date) > strtotime($current_date))
{
echo "Future date not accepted";
}
Hope this helps.

PHP - checking if two dates match but ignoring the year

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()))) )

Optional month or day in MySQL date field from PHP

I have a problem where I need to handle dates where the month and day parts are optional. For example, the year will always be known but sometimes the day or month and day will be unknown.
In MySQL I can create a table with a date field and while I can't find any reference in the MySQL Manual it will accept the following as valid:
(YYYY-MM-DD format):
2011-02-10 // Current date
2011-02-00 // Day unknown so replaced with 00
2011-00-00 // Day and month unkown so replaced with 00-00
Test calculations from within the database work fine so I can still sort results easily. In the manual it says that month needs to be between 01 and 12, and day between 01 and 31 - but it does accept 00.
First question: Am I going to run into trouble using 00 in the month or day parts or is this perfectly acceptable?
Next question: Is there a PHP function (or MySQL format command) that will automatically format the following dates into the required format string?
2011 becomes 2011-00-00
2011-02 becomes 2011-02-00
Or do I need write a special function to handle this?
The following doesn't work:
<?php
$date = date_create_from_format('Y-m-d', '2011-00-00');
echo date_format($date, 'Y-m-d');
// Returns 2010-11-30
$date = date_create_from_format('Y-m-d', '2011-02-00');
echo date_format($date, 'Y-m-d');
// Returns 2011-01-31
?>
Third question: Is there a PHP function (or MySQL command) to format the dates for use in PHP?
Finally, is this the best approach? Or is there a 'best practise' method?
EDIT:
Here is what I'm currently doing:
A date field can accept a date in the format YYYY, YYYY-MM, or YYYY-MM-DD and before sending to the database it is processed in this function:
/**
* Takes a date string in the form:
* YYYY or
* YYYY-MM or
* YYYY-MM-DD
* and validates it
*
* Use date_format($date, $format); to reverse.
*
* #param string $phpDate Date format [YYYY | YYYY-MM | YYYY-MM-DD]
*
* #return array 'date' as YYYY-MM-DD, 'format' as ['Y' | 'Y-m' | 'Y-m-d'] or returns false if invalid
*/
function date_php2mysql($phpDate) {
$dateArr = false;
// Pattern match
if (preg_match('%^(?P<year>\d{4})[- _/.]?(?P<month>\d{0,2})[- _/.]?(?P<day>\d{0,2})%im', trim($phpDate), $parts)) {
if (empty($parts['month'])) {
// Only year valid
$date = $parts['year']."-01-01";
$format = "Y";
} elseif (empty($parts['day'])) {
// Year and month valid
$date = $parts['year']."-".$parts['month']."-01";
$format = "Y-m";
} else {
// Year month and day valid
$date = $parts['year']."-".$parts['month']."-".$parts['day'];
$format = "Y-m-d";
}
// Double check that it is a valid date
if (strtotime($date)) {
// Valid date and format
$dateArr = array('date' => $date, 'format' => $format);
}
} else {
// Didn't match
// Maybe it is still a valid date
if (($timestamp = strtotime($phpDate)) !== false) {
$dateArr = array('date' => date('Y-m-d', $timestamp), 'format' => "Y-m-d");
}
}
// Return result
return $dateArr;
}
So it pattern matches the input $phpDate where it must begin with 4 digits, then optionally pairs of digits for the month and the day. These are stored in an array called $parts.
It then checks if months or days exist, specifying the format string and creating the date.
Finally, if everything checks out, it returns a valid date as well as a format string. Otherwise it returns FALSE.
I end up with a valid date format for my database and I have a way of using it again when it comes back out.
Anyone think of a better way to do this?
I have a problem where I need to handle dates where the month and day parts are optional.
For example, the year will always be known but sometimes the day or month and day will be
unknown.
In many occasions, we do need such 'more or less precise' dates, and I use such dates as 2011-04-01 (precise), as well as 2011-04 (= April 2011) and 2011 (year-only date) in archives metadata. As you mention it, MySQL date field tolerates '2011-00-00' though no FAQs tell about it, and it's fine.
But then, I had to interface the MySQL database via ODBC and the date fields
are correctly translated, except the 'tolerated' dates (Ex: '2011-04-00' results empty in the resulting MySQL-ODBC-connected ACCESS database.
For that reason, I came to the conclusion that the MySQL date field could be converted in a plain VARCHAR(10) field : As long as we don't need specific MySQL date functions, it works fine, and of course, we can still use php date functions and your fine date_php2mysql() function.
I would say that the only case when a MySQL date field is needed
is when one needs complex SQL queries, using MySQL date functions in the query itself.
(But such queries would not work anymore on 'more or less precise' dates!...)
Conclusion : For 'more or less precise' dates,
I presently discard MySQL date field and use plain VARCHAR(10) field
with aaaa-mm-jj formated data. Simple is beautiful.
Since the data parts are all optional, would it be tedious to store the month, day, and year portions in separate integer fields? Or in a VARCHAR field? 2011-02-00 is not a valid date, and I wouldnt't think mysql or PHP would be excited about it. Test it out with str_to_time and see what kind of results you get, also, did you verify that the sorting worked right in MySQL? If the docs say that 1 through 31 is required, and it is taking 00, you might be relying on what is, in essence, a bug.
Since 2011-02-00 is not a valid date, none of PHP's formatting functions will give you this result. If it handled it at all, I wouldn't be surprised if you got 2001-01-31 if you tried. All the more reason to either store it as a string in the database, or put the month, day, and year in separate integer fields. If you went with the latter route, you could still do sorting on those columns.
I have also encountered this problem. I ended up using the PEAR Date package. Most date classes won't work with optional months or optional days, but the PEAR Date package does. This also means you don't need custom formatting functions and can use the fancy formatting methods provided by the Date package.
I have found this link in a textbook. This states that month and day values can be zero to allow for the possiblity of storing incomplete or unknown data
http://books.google.co.uk/books?id=s_87mv-Eo4AC&pg=PA145&lpg=PA145&dq=mysql+date+of+death+when+month+unknown&source=bl&ots=tcRGz3UDtg&sig=YkwpkAlDtBP1KKTDtqSyZCl63hs&hl=en&ei=Btf5TbL1NIexhAfkveyTAw&sa=X&oi=book_result&ct=result&resnum=8&ved=0CFMQ6AEwBw#v=onepage&q&f=false
If you pull your date in pieces from the database you can get it as if it's 3 fields.
YEAR(dateField) as Year, MONTH(dateField) as Month, DAY(dateField) as DAY
Then pushing those into the corresponding fields in the next bit of PHP will give you the result you're looking for.
$day = 0;
$month = 0;
$year = 2013;
echo $datestring;
$format = "Y";
if($month)
{
$format .= "-m";
if($day)
$format .="-d";
else
$day = 1;
}
else
{
$month = 1;
$day = 1;
}
$datestring = strval($year)."-".strval($month)."-".strval($day);
$date = date($format, strtotime($datestring));
echo $date; // "2013", if $month = 1, "2013-01", if $day and $month = 1, "2013-01-01"

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