Adding one day to a date

Question

My code to add one day to a date returns a date before day adding   2009-09-30 20 24 00 date after adding one day SHOULD be rolled over to the next month  1970-01-01 17 33 29   lt  php        add day to date test for month roll over       stop date   date  Y-m-d H i s   strtotime  2009-09-30 20 24 00          echo  date before day adding     stop date         stop date   date  Y-m-d H i s   strtotime   1 day    stop date         echo   date after adding one day  SHOULD be rolled over to the next month     stop date    gt    I have used pretty similar code before  what am I doing wrong here

User · Answer

lt  php  function plusTimetoOldtime  Old Time  getFormat  Plus Time        return date  getFormat strtotime date  getFormat  Old Time   Plus Time        Old Time   strtotime  quot now quot     Plus Time     1 day    getFormat    Y-m-d H i s    echo plusTimetoOldtime  Old Time  getFormat  Plus Time      gt

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lt  php  stop date    2009-09-30 20 24 00   echo  date before day adding       stop date    stop date   date  Y-m-d H i s   strtotime  stop date      1 day     echo  date after adding 1 day       stop date    gt    For PHP 5 2 0   you may also do as follows    stop date   new DateTime  2009-09-30 20 24 00    echo  date before day adding       stop date- gt format  Y-m-d H i s      stop date- gt modify   1 day    echo  date after adding 1 day       stop date- gt format  Y-m-d H i s

User · Answer

Simple to read and understand way    original date    2009-09-29     time original   strtotime  original date    time add         time original    3600 24     add seconds of one day   new date        date  Y-m-d    time add    echo  new date

User · Answer

I always just add 86400  seconds in a day     stop date   date  Y-m-d H i s   strtotime  2009-09-30 20 24 00     86400    echo  date after adding 1 day     stop date     It s not the slickest way you could probably do it  but it works

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date   new DateTime  2000-12-31      date- gt modify   1 day    echo  date- gt format  Y-m-d       n

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Since you already have an answer to what s wrong with your code  I can bring another perspective on how you can play with datetimes generally  and solve your problem specifically   Oftentimes you find yourself posing a problem in terms of solution  This is just one of the reasons you end up with an imperative code  It s great if it works though  there are just other  arguably more maintainable alternatives  One of them is a declarative code  The point is asking what you need  instead of how to get there   In your particular case  this can look like the following  First  you need to find out what is it that you re looking for  that is  discover abstractions  In your case  it looks like you need a date  Not just any date  but the one having some standard representation  Say  ISO8601 date  There are at least two implementations  the first one is a date parsed from an ISO8601-formatted string  or a string in any other format actually   and the second is some future date which is a day later  Thus  the whole code could look like that    new Future      new DateTimeParsedFromISO8601  2009-09-30 20 24 00        new OneDay          - gt value      For more examples with datetime juggling check out this one

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Simplest solution    date   new DateTime   1 day    echo  date- gt format  Y-m-d H i s

User · Answer

It Worked for me  For Current Date   date   date  Y-m-d   strtotime   1 day       for anydate   date  Y-m-d   strtotime   1 day   strtotime  date

User · Answer

Try this  echo date  Y-m-d H i s  date strtotime   1 day   strtotime  2009-09-30 20 24 00

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The following code get the first day of January of current year  but it can be a another date  and add 365 days to that day  but it can be N number of days  using DateTime class and its method modify   and format     echo  new DateTime  new DateTime   - gt modify  first day of January this year  - gt format  Y-m-d    - gt modify   365 days  - gt format  Y-m-d

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While I agree with Doug Hays  answer  I ll chime in here to say that the reason your code doesn t work is because strtotime   expects an INT as the 2nd argument  not a string  even one that represents a date   If you turn on max error reporting you ll see this as a  A non well formed numeric value  error which is E NOTICE level

User · Answer

The modify   method that can be used to add increments to an existing DateTime value   Create a new DateTime object with the current date and time    due dt   new DateTime      Once you have the DateTime object  you can manipulate its value by adding or subtracting time periods    due dt- gt modify   1 day      You can read more on the PHP Manual

[php] Adding one day to a date

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