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If I have two dates 20-4-2010 and 22-4-2010 in two text box and I want the dates to be like this 20, 21, 22. How do I get that ?



I am pretty sure this has been answered a quadrillion times before, but anyway:

$start = strtotime('20-04-2010 10:00');
$end   = strtotime('22-04-2010 10:00');
for($current = $start; $current <= $end; $current += 86400) {
    echo date('d-m-Y', $current);

The 10:00 part is to prevent the code to skip or repeat a day due to daylight saving time.

By giving the number of days:

for($i = 0; $i <= 2; $i++) {
    echo date('d-m-Y', strtotime("20-04-2010 +$i days"));

With PHP5.3

$period = new DatePeriod(
    new DateTime('20-04-2010'),
    DateInterval::createFromDateString('+1 day'),
    new DateTime('23-04-2010') // or pass in just the no of days: 2

foreach ( $period as $dt ) {
  echo $dt->format( 'd-m-Y' );
Thursday, October 6, 2022

Here's what worked best for me when trying to script this (in case anyone else comes across this like I did):

$ pecl -d php_suffix=5.6 install <package>
$ pecl uninstall -r <package>

$ pecl -d php_suffix=7.0 install <package>
$ pecl uninstall -r <package>

$ pecl -d php_suffix=7.1 install <package>
$ pecl uninstall -r <package>

The -d php_suffix=<version> piece allows you to set config values at run time vs pre-setting them with pecl config-set. The uninstall -r bit does not actually uninstall it (from the docs):

vagrant@homestead:~$ pecl help uninstall
pecl uninstall [options] [channel/]<package> ...
Uninstalls one or more PEAR packages.  More than one package may be
specified at once.  Prefix with channel name to uninstall from a
channel not in your default channel (

  -r, --register-only
        do not remove files, only register the packages as not installed

The uninstall line is necessary otherwise installing it will remove any previously installed version, even if it was for a different PHP version (ex: Installing an extension for PHP 7.0 would remove the 5.6 version if the package was still registered as installed).

Monday, December 12, 2022

Update 2018-04-20: it seems that OP @Joshkunz was asking for finding which months are between two dates, instead of "how many months" are between two dates. So I am not sure why @JohnLaRooy is upvoted for more than 100 times. @Joshkunz indicated in the comment under the original question he wanted the actual dates [or the months], instead of finding the total number of months.

So it appeared the question wanted, for between two dates 2018-04-11 to 2018-06-01

Apr 2018, May 2018, June 2018 

And what if it is between 2014-04-11 to 2018-06-01? Then the answer would be

Apr 2014, May 2014, ..., Dec 2014, Jan 2015, ..., Jan 2018, ..., June 2018

So that's why I had the following pseudo code many years ago. It merely suggested using the two months as end points and loop through them, incrementing by one month at a time. @Joshkunz mentioned he wanted the "months" and he also mentioned he wanted the "dates", without knowing exactly, it was difficult to write the exact code, but the idea is to use one simple loop to loop through the end points, and incrementing one month at a time.

The answer 8 years ago in 2010:

If adding by a week, then it will approximately do work 4.35 times the work as needed. Why not just:

1. get start date in array of integer, set it to i: [2008, 3, 12], 
       and change it to [2008, 3, 1]
2. get end date in array: [2010, 10, 26]
3. add the date to your result by parsing i
       increment the month in i
       if month is >= 13, then set it to 1, and increment the year by 1
   until either the year in i is > year in end_date, 
           or (year in i == year in end_date and month in i > month in end_date)

just pseduo code for now, haven't tested, but i think the idea along the same line will work.

Saturday, September 10, 2022

I think there is no out-of-the-box API to provide the result in the format you mentioned. You need to use the DATEDIFF function to get the difference in the least denomination you need and then divide the result with appropriate value to get the duration in the format required. Something like this:

DECLARE @duration INT

SELECT @start = '2009-10-06', @end = '2011-07-15'
SELECT @duration = DATEDIFF(mm, @start, @end)
SELECT CONVERT(NVARCHAR, @duration / 12) + '.' + CONVERT(NVARCHAR, @duration % 12)

This can be better achieved by writing a function that would take the dates and least denomination and returns the duration in the format needed, as it would require TSQL and plain SQL wouldn't suffice.

Tuesday, November 8, 2022

Never used any of those, but they look interesting..

Take a look at Gearman as well.. more overhead in systems like these but you get other cool stuff :) Guess it depends on your needs ..

Friday, November 11, 2022
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