Add number of days to a date

问题:

I want to add number of days to current date: I am using following code:

$i=30;
echo $date = strtotime(date(\"Y-m-d\", strtotime($date)) . \" +\".$i.\"days\");

But instead of getting proper date i am getting this: 2592000

Please suggest.

回答1:

This should be

echo date(\'Y-m-d\', strtotime(\"+30 days\"));

strtotime

expects to be given a string containing a US English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.

while date

Returns a string formatted according to the given format string using the given integer timestamp or the current time if no timestamp is given.

See the manual pages for

• http://www.php.net/manual/en/function.strtotime.php
• http://www.php.net/manual/en/function.date.php

and their function signatures.

回答2:

This one might be good

function addDayswithdate($date,$days){

    $date = strtotime(\"+\".$days.\" days\", strtotime($date));
    return  date(\"Y-m-d\", $date);

}

回答3:

$date = new DateTime();
$date->modify(\'+1 week\');
print $date->format(\'Y-m-d H:i:s\');

or print date(\'Y-m-d H:i:s\', mktime(date(\"H\"), date(\"i\"), date(\"s\"), date(\"m\"), date(\"d\") + 7, date(\"Y\"));

回答4:

$today=date(\'d-m-Y\');
$next_date= date(\'d-m-Y\', strtotime($today. \' + 90 days\'));
echo $next_date;

回答5:

You can add like this as well, if you want the date 5 days from a specific date :

You have a variable with a date like this (gotten from an input or DB or just hard coded):

$today = \"2015-06-15\"; // Or can put $today = date (\"Y-m-d\");

$fiveDays = date (\"Y-m-d\", strtotime ($today .\"+5 days\"));

echo $fiveDays; // Will output 2015-06-20

回答6:

Keep in mind, the change of clock changes because of daylight saving time might give you some problems when only calculating the days.

Here\'s a little php function which takes care of that:

function add_days($date, $days) {
    $timeStamp = strtotime(date(\'Y-m-d\',$date));
    $timeStamp+= 24 * 60 * 60 * $days;

    // ...clock change....
    if (date(\"I\",$timeStamp) != date(\"I\",$date)) {
        if (date(\"I\",$date)==\"1\") { 
            // summer to winter, add an hour
            $timeStamp+= 60 * 60; 
        } else {
            // summer to winter, deduct an hour
            $timeStamp-= 60 * 60;           
        } // if
    } // if
    $cur_dat = mktime(0, 0, 0, 
                      date(\"n\", $timeStamp), 
                      date(\"j\", $timeStamp), 
                      date(\"Y\", $timeStamp)
                     ); 
    return $cur_dat;
}

回答7:

You could also try:

$date->modify(\"+30 days\");

回答8:

You can do it by manipulating the timecode or by using strtotime(). Here\'s an example using strtotime.

$data[\'created\'] = date(\'Y-m-d H:i:s\', strtotime(\"+1 week\"));

回答9:

You can use strtotime()
$data[\'created\'] = date(\'Y-m-d H:m:s\', strtotime(\'+1 week\'));

回答10:

I know this is an old question, but for PHP <5.3 you could try this:

$date = \'05/07/2013\';
$add_days = 7;
$date = date(\'Y-m-d\',strtotime($date) + (24*3600*$add_days)); //my preferred method
//or
$date = date(\'Y-m-d\',strtotime($date.\' +\'.$add_days.\' days\');

回答11:

You could use the DateTime class built in PHP. It has a method called \"add\", and how it is used is thoroughly demonstrated in the manual: http://www.php.net/manual/en/datetime.add.php

It however requires PHP 5.3.0.

回答12:

$date = \"04/28/2013 07:30:00\";

$dates = explode(\" \",$date);

$date = strtotime($dates[0]);

$date = strtotime(\"+6 days\", $date);

echo date(\'m/d/Y\', $date).\" \".$dates[1];

回答13:

You may try this.

$i=30;
echo  date(\"Y-m-d\",mktime(0,0,0,date(\'m\'),date(\'d\')+$i,date(\'Y\')));

回答14:

Simple and Best

echo date(\'Y-m-d H:i:s\').\"\\n\";
echo \"<br>\";
echo date(\'Y-m-d H:i:s\', mktime(date(\'H\'),date(\'i\'),date(\'s\'), date(\'m\'),date(\'d\')+30,date(\'Y\'))).\"\\n\";

Try this

回答15:

//add the two day

$date = \"**2-4-2016**\"; //stored into date to variable

echo date(\"d-m-Y\",strtotime($date.**\' +2 days\'**));

//print output
**4-4-2016**

回答16:

Use this addDate() function to add or subtract days, month or years (you will need the auxiliar function reformatDate() as well)

/**
 * $date self explanatory
 * $diff the difference to add or subtract: e.g. \'2 days\' or \'-1 month\'
 * $format the format for $date
 **/
    function addDate($date = \'\', $diff = \'\', $format = \"d/m/Y\") {
        if (empty($date) || empty($diff))
            return false;
        $formatedDate = reformatDate($date, $format, $to_format = \'Y-m-d H:i:s\');
        $newdate = strtotime($diff, strtotime($formatedDate));
        return date($format, $newdate);
    }
    //Aux function
    function reformatDate($date, $from_format = \'d/m/Y\', $to_format = \'Y-m-d\') {
        $date_aux = date_create_from_format($from_format, $date);
        return date_format($date_aux,$to_format);
    }

Note: only for php >=5.3

回答17:

Use the following code.

<?php echo date(\'Y-m-d\', strtotime(\' + 5 days\')); ?>

Reference has found from here - How to Add Days to Current Date in PHP

回答18:

//Set time zone
date_default_timezone_set(\"asia/kolkata\");
$pastdate=\'2016-07-20\';
$addYear=1;
$addMonth=3;
$addWeek=2;
$addDays=5;
$newdate=date(\'Y-m-d\', strtotime($pastdate.\' +\'.$addYear.\' years +\'.$addMonth. \' months +\'.$addWeek.\' weeks +\'.$addDays.\' days\'));
echo $newdate;

回答19:

Even though this is an old question, this way of doing it would take of many situations and seems to be robust. You need to have PHP 5.3.0 or above.

$EndDateTime = DateTime::createFromFormat(\'d/m/Y\', \"16/07/2017\");
$EndDateTime->modify(\'+6 days\');
echo $EndDateTime->format(\'d/m/Y\');

You can have any type of format for the date string and this would work.