Assume that i have two variables in php:
$year
$month
Then I want to make another variable:
$date
which:
$date=$year-$month-25
So, if I have 2012 in $year and 7 for $month, $date will be 2012-07-25.
Actually, I will compare it with some date in MySQL.
$year and $month are inputted by user.
anybody have a solution?
The solution either how to make $date or anything as long it can be comparred with a date in mysql.
Thanks before. ^^
You can make a unix timestamp through this:
$myDate = mktime(0, 0, 0, $month, 25, $year);
This is a pretty useful thing to have, as you can format it into all sorts of nice via:
echo date("Y-m-d", $myDate);
// Prints something like: 2012-07-25
or
echo date("l", $myDate);
// Prints something like: Monday
or
date('l jS \of F Y h:i:s A', $myDate);
// Prints something like: Monday 8th of August 2005 03:12:46 PM
You can do as follows
$complete_date = $year."-".$month."-25";
which gives you 2012-7-25
Please, read the "php manual" for concat your PHP string.
it's not
$date = $year-$month-25;
it is
$date = $year . '-' . $month . '- 25';
or
$date = $year . "-" . $month . "- 25";
but simple quote is more optimize for php string.
The solution either how to make $date or anything as long it can be
comparred with a date in mysql
The key here is use of strtotime to create and compare.
MySQL dates can be converted to integer through the use of strototime:
strtotime($mysql_date);
Then you can get time() and compare to two:
time()<>strtotime($mysql_date) // then the two dates are not equal.
You can use mktime function
$date = date('Y-m-d',mktime(0,0,0,$month,25,$year));
Well, I would use mktime to get the timestamp of the date ( http://php.net/manual/de/function.mktime.php ) and use the command unix_timestamp(date(yourfield)) in mysql to compare them.
(the date() withing unix_timestamp is only required when you save datetime values and not pure date values)
Since mysql dates are usually in this format Y-m-d by default, you can use $thedate = date('Y-m-d',mktime(0,0,0,$month,25,$year)); where $month and $year are based on the user input. Of course you have to make the user input it in the format you want by using select/lists.
Related
I'm trying to extract just the day of the month from the dateTime formatted as follows:
$today = date('Y-m-d\TH:i:sP', strtotime('today'));
But, when I try extracting just the day of the month, I get '31'.
I'm using: $day = date('d', $today);
Which, I'm guessing, is incorrect.
This is because the second parameter to date() needs to be a Unix Timestamp. You're passing it string. As a result you get a date of Dec 31, 1969.
All of that code is unnecessary anyways as all you need is:
$day = day('d');
If you're going to only have access to the date string you must convert it to Unix Timestamp before passing it to date().
To extract days (or other parts of a datetime), I use the format function:
$datetime = new DateTime('2000-01-10', new DateTimeZone('Pacific/Nauru'));
$day = $datetime->format('d');
echo $day;
Use any format form the PHP Manual.
Hope this will solve your problem
$today = date('Y-m-d\TH:i:sP', strtotime('today'));
$day = date('d', strtotime($today)); // here is the difference,
// instead of $today use strtotime($today)
I know the year and the day of year (0-365).
Is there any way to get back from it to timestamp?
Something like:
$day=date('z');
$year=2012;
$timestamp=strtotime($day.'-nd day of '.$year);
Try the following:
$timestamp = mktime(0,0,0,1,$day,$year);
It will produce a timestamp where the time is set to 0:00:00.
(I am not sure if it will behave correctly with respect to daylight savings though...)
Check the PHP manual for further details: PHP: mktime - Manual
Maybe strtotime gets handy here:
php -r 'echo #date("Y-m-d H:s", strtotime("january 2012 +200 day")).PHP_EOL;'
In your example $timestamp = strtotime("january $year +$day day");
Try using createFromFormat :
$date = DateTime::createFromFormat('z Y', $day . ' ' . $year);
echo $date->getTimestamp();
I have a month value (1-12), day value (1-31), and a year value (2010,2011,2012). I also have a hour value and a minute value.
How can I give this to strtotime() in a way it can convert it to a timestamp?
why convert string to date when you already know year month and date.
use setDate funtion
<?php
$date = new DateTime();
$date->setDate(2001, 2, 3);
echo $date->format('Y-m-d');
?>
Given the variables $year $month $day $hour $minute you can do:
strtotime("$year-$month-$day $hour:$minute");
Be careful to enclose the variables with ", never in this case with '.
UPDATE (thanks to comments of #Clockwork and #Tadeck):
In case there is a variable $timeofday that represents the time of day (i.e. AM or PM),
then you can parse it this with:
strtotime("$year-$month-$day $hour:$minute$timeofday");
that is to say, just attach that variable to the end of the text.
Is strtotime the best tool for this job? What about mktime()?
$time = mktime($hour, $minute, 0, $month, $day, $year);
You can provide it to function strtotime() in many ways, as mentioned in documentation. Some examples include:
$your_time = strtotime('12/31/2011 9:59');
$your_time = strtotime('2011-12-31 9:59');
$your_time = strtotime('December 31, 2011 9:59');
etc. It really is very flexible.
You can find the list of valid formats in the documentation, and that is (from the "Compound Formats" list in the mentioned documentation) for example:
10/Oct/2000:13:55:36 -0700,
2008:08:07 18:11:31,
2008-08-07 18:11:31,
2008-07-01T22:35:17.02,
2008-07-01T22:35:17.03+08:00,
20080701T22:38:07,
20080701T9:38:07,
20080701t223807,
20080701T093807,
2008-7-1T9:3:37,
(this is really copy of the documentation)
Use it like this strtotime("YYYY-mm-DD HH:MM AM/PM"):
echo date("d F Y h:i:s A", strtotime("2011-06-01 11:15 PM")) . "\n";
OUTPUT
01 June 2011 11:15:00 PM
Y-m-d hh:mm will work
echo strtotime('2011-12-14 11:44 am');
cit #Pekka :)
strtotime($month."-".$day."-".$year)
I am pulling the dates of various posts from a database. The dates are in the following format:
2009-08-12
Numeric Year - Numeric Month - Numeric Day
How can I reformat these dates to something more user friendly like:
August 12, 2009
Numeric Month Numeric Date, Numeric Year
Assuming that the date gotten from the mysql database is stored in a variable called:
$date = $row['date_selected'];
Unlike the strtotime based examples, this allows you to ensure the month and day are interpreted in the correct order regardless of locale settings specified on the server.
$date = DateTime::createFromFormat('Y-m-d', '2009-08-12');
$output = $date->format('F j, Y');
date("F d, Y", strtotime($input))
$new_format = date("Your Date String", strtotime($date));
See:
- http://php.net/strtotime
- http://php.net/date
Basically, if strtotime() can read it correctly, you can reformat it anyway you please.
In this case, Year - Month - Day is a properly recognized strtotime() format, this might not be the case for other formats.
You might consider doing your date formatting in MySQL with your select statement:
DATE_FORMAT(date,'%M %e, %Y') as date_selected
http://www.w3schools.com/sql/func_date_format.asp
<?php
echo date('F j, Y', strtotime($date));
You might want to look at the php function strtotime:
http://php.net/manual/en/function.strtotime.php
It'll parse a large number of date representations to a Unix timestamp.
Then use the date function.
Using strtodate or explode to split the date into its different components, you can then use the date function with the appropriate format string:http://php.net/manual/en/function.date.php
$date = "2009-08-12";
list($year,$month,$day) = explode("-",$date);
$formattedDate = date("F d, Y", mktime(0,0,0,$month,$day,$year));
Outputs: "August 12, 2009"
<?php
//Date Formatter
/*
date: date you want to convert
format: its current format ie m-d-Y, m/d/Y, Y-m-d, Y/m/d
delimS: Current delimiter ie - or / or .
delimF: The delimiter you want for the result
NOTE: this will only convert m-d-Y to Y-m-d and back
*/
function dtform($date,$format,$delimS,$delimF){
$dateFinal = '';
if($format == 'm'.$delimS.'d'.$delimS.'Y'){
$dateFinal_exp = explode($delimS,$date);
$dateFinal = $dateFinal_exp[2].$delimF.$dateFinal_exp[0].$delimF.$dateFinal_exp[1];
}else if($format == 'Y'.$delimS.'m'.$delimS.'d'){
$dateFinal_exp = explode($delimS,$date);
$dateFinal = $dateFinal_exp[1].$delimF.$dateFinal_exp[2].$delimF.$dateFinal_exp[0];
}
return $dateFinal;
}
?>
Use it like this:
// February 1, 2005
print date ("F j, Y", mktime (0,0,0,14,1,2004));
I'm trying to convert 2010-02 to February, 2010. But, I keep getting December, 1969
I've tried using mktime, strtotime, and some combination of the two, but still haven't been able to do it...
This is what I tried most recently...
$path_title = date('F, Y', mktime(0,0,0,2,0,2010));
This would be a way to do it:
$dateString = '2010-02';
list($year, $month) = explode('-', $dateString);
$timeStamp = mktime(0, 0, 0, $month, 1, $year);
echo date('F, Y', $timestamp);
Another way would be:
$dateString = '2010-02';
$timestamp = strtotime($dateString . '-01');
echo date('F, Y', $timestamp);
strtotime can't handle ambiguous dates like "2010-02", but if you make it a full date it should work.
Otherwise, you may want to look into something like DateTime::createFromFormat.
Try this:
$str = '2010-02';
echo date('F, Y',mktime(0,0,0,substr($str,-2),1,substr($str,0,4)));
You have to make sure you use valid values to mktime(). In your example that you edited into the question, you have 0 as the day, which is effectively the first day minus one, which puts you into the previous month.