How can I find out the date of the day before a da

问题:

I have a system that compares fiscal year up to the current date to the same date range of the year before. You can adjust the year and month to look at and it always compares the same date range to the year previous. I have it set so if the current day is leap day it compares to the 28th of last year, and if last year was leap year and today is feb 28th it compares to last year up to 29th. if you look at a month other than the current month it shows up to the last day of that month, otherwise up to current date.

OK that works fine right now, but now my employers don't want it to be up to the current date they want it to go up to yesterdays date. How can I do that, my main concerns are what if today is the 1st of the month, or what if today is the first day of the fiscal year.

Here is the code I have now:

function create_YTD_XML()
{
    global $month;
    global $year;

    $last_year = $year - 1;

    if($year == date('Y') && $month == date('m'))
    {
        $this_day = date('d');
    }
    else
    {
        $this_day = date('t', mktime(0, 0, 0, $month, 1, $year)); // LAST DAY OF MONTH
    }

    if(is_leap_year($year) && $this_day == 29)
    {
        $last_day = 28;
    }
    else if(is_leap_year($last_year) && $this_day == 28)
    {
        $last_day = 29;
    }
    else
    {
        $last_day = $this_day;
    }

    if($month >= 2)
    {
        $this_year_start = $year;
        $last_year_start = $last_year;
    }
    else
    {
        $this_year_start = $year - 1;
        $last_year_start = $last_year - 1;
    }

    $this_ytd_start = $this_year_start.'-02-01';
    $last_ytd_start = $last_year_start.'-02-01';

    $this_ytd_end = $year.'-'.str_pad($month, 2, "0", STR_PAD_LEFT).'-'.$this_day;
    $last_ytd_end = $last_year.'-'.str_pad($month, 2, "0", STR_PAD_LEFT).'-'.$last_day;


}

What would be the best solution?

Thanks!

回答1:

strtotime() will do the trick. Convert your previous date to a Unix timestamp using mktime(), then use it like this:

$from_unix_time = mktime(0, 0, 0, $month, $day, $year);
$day_before = strtotime("yesterday", $from_unix_time);
$formatted = date('Y-m-d', $day_before);

回答2:

You can also use strtotime function using words like this:

$date = '2012-11-08';
$day_before = date( 'Y-m-d', strtotime( $date . ' -1 day' ) );

The output of $day_before:

2012-11-07

回答3:

You can rely on mktime()’s ability to work with “incorrect” values. Let’s assume $year, $month and $day are current day.

$yesterday = mktime (0, 0, 0, $month, $day - 1, $year);
echo date ('Y-m-d', $yesterday);

Don’t worry, it will just do the right thing: try yourself with different values.

回答4:

Note: The following answer does not take daylight savings time into account; during certain times of the year, it will return an inaccurate result. See the comments for more information.

---

Since the length of a day is constant (86,400 seconds), you can use arithmetic to determine the timestamp of the day before a given date:

$yesterday = date('Y-m-d', strtotime($current_date) - 86400);

$current_date is the date you want to find the day before. If it is already a unix timestamp, you can dispense with strtotime() entirely:

$yesterday = date('Y-m-d', $current_date_as_unix_timestamp - 86400);

回答5:

$date = date('Y-m-d',strtotime('yesterday'));

回答6:

I would use strtotime indeed, in this way:

$date="2014-02-20";
echo date("Y-m-d",strtotime($date." -1 day"));

will output:

2014-02-19

回答7:

I think it depends on the behavior your employer wants. There's no "right" answer—if today is the start of a fiscal year, do they want to see the last two fiscal years compared (which is what would happen if you used yesterday as the "up to" date) or do they want to see day 1 of this fiscal year compared to day 1 of last? You need to sit down and figure out the potential edge cases and then talk with your employer about what they want.

回答8:

You can also use following code its works:

$current_date = strtotime('25-11-2012');
echo $yesterday = date('Y-m-d', strtotime('-1 day', $current_date));

And output of code is following:

2012-11-24

回答9:

One day before time at fixed 07:59:59

$demo = date("Y-m-d H:i:s", mktime(7, 59, 59, date("m"),date("d")-1,date("Y")));