date of Yesterday

ctrohaya · #1 28-Nov-2006, 03:48

I want to create the date of yesterday for date entered. I got the date for yesterday as:

PHP Code:


$yesterday  = mktime (date ("H"), date ("i"), date ("s"), date("m"), date "d") -1, date("Y"));
    $bhari = date ("d", $yesterday);
    $bln = date ("m", $yesterday);
    $tahun = date ("Y", $yesterday);

but, if I enter day=6, month=10 year-2006, how can I get yesterday's date for above date? And how can I get the date for day before $yesterday?

JdS · #2 28-Nov-2006, 07:26

You can easily get the date of yesterday by using strtotime().

Here are a couple of examples:

PHP Code:


<?php
// yesterday in 'd-m-Y' format
$uts['yesterday'] = strtotime( '-1 days' ); // or, strtotime( 'yesterday' );
echo "Yesterday: ", date( 'd-m-Y', $uts['yesterday'] ), "<br />\n";

// day-before-yesterday in 'd-m-Y' format
$uts['day-before-yesterday'] = strtotime( '-2 days' );
echo "The day before yesterday: ", date( 'd-m-Y', $uts['day-before-yesterday'] ), "<br />\n";
?>

You can refer to http://www.gnu.org/software/tar/manu...19.html#SEC119 for examples of valid (relative) date strings that you can use with strtotime().

ctrohaya · #3 28-Nov-2006, 21:32

It uses the system date, but if I enter my own date for example: day=6, month = 10, year = 2006, how can I get the date for day before above date.

JdS · #4 29-Nov-2006, 01:57

If you look at the details of the strtotime function, you will see that it expects an optional timestamp value too.

int strtotime ( string time [, int now] )

So, if you omit passing a timestamp in the second parameter, you will be expecting the results relative to the system date, as per my examples above.

If you have your own timestamp, you can submit it to this second parameter and the calculations will be relative to this value, and not the system date.

Here is a new example:

PHP Code:


<?php
$mydate = '2006-10-06'; // ISO 8601 format.
                        // see http://www.gnu.org/software/tar/manual/html_node/tar_115.html#SEC115
                        // for other valid date/calendar formats.

// mydate into unixtimestamp
$uts['mydate'] = strtotime( $mydate );

// mydate - 1 (days)
$uts['mydate-1d'] = strtotime( '-1 day', $uts['mydate'] );

// mydate - 2 (days)
$uts['mydate-2d'] = strtotime( '-2 days', $uts['mydate'] );

// mydate - 1 (months)
$uts['mydate-1m'] = strtotime( '-1 months', $uts['mydate'] );

// display results.
echo "mydate: ",                date( 'd-m-Y', $uts['mydate'] ),    "<br />\n";
echo "mydate minus 1 day: ",    date( 'd-m-Y', $uts['mydate-1d'] ),    "<br />\n";
echo "mydate minus 2 days: ",    date( 'd-m-Y', $uts['mydate-2d'] ),    "<br />\n";
echo "mydate minus 1 month: ",    date( 'd-m-Y', $uts['mydate-1m'] ),    "<br />\n";
?>

admin · #5 29-Nov-2006, 03:10

Yes, of course.

$mydate = '2006-10-01' is just a string.

It could have been written like this as well: $mydate = "$year-$month-$day"; where $year is 2006, $month is 10 and $day is 6.

ctrohaya · #6 29-Nov-2006, 03:29

Thank you so much, I've got the solution.

admin · #7 29-Nov-2006, 03:37

One last thing, reading the page again, I found this important detail.

Quote:

For numeric months, the ISO 8601 format `year-month-day' is allowed, where year is any positive number, month is a number between 01 and 12, and day is a number between 01 and 31. A leading zero must be present if a number is less than ten. If year is 68 or smaller, then 2000 is added to it; otherwise, if year is less than 100, then 1900 is added to it. The construct `month/day/year', popular in the United States, is accepted. Also `month/day', omitting the year.

Emphasis mine.

So, '2006-10-01' is correct and '2006-10-1' is not!?

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