i want to separate date, month, time, hour, min ,sec getting from database, for count down timer,
in data base stored date is 2010/10/10:
me using this code :
$m=date('m',$row['Date']);
$d=date('d',$row['Date']);
output is month=31
date = 12....
If you want them separetely, you can use the explode function with list eg:
list($year, $month, $day) = explode('/', $row['Date']);
Also, you can use the strtotime function for getting individual values:
echo 'Day' . date('d', strtotime($row['Date']));
echo 'Month' . date('m', strtotime($row['Date']));
echo 'Year' . date('Y', strtotime($row['Date']));
$d = // Pull your date from the db
$date = new DateTime($d);
$day = $date->format('l'); // Creates day name
$day = $date->format('j') // Creates date number
$month = $date->format('n'); // Creates month number
$year = $date->format('Y'); // Creates year
..and so forth.
You can find formatting options here, http://php.net/manual/en/function.date.php
Related
I have three pieces of information extracted from my database, the year, the number of week (1-52), and the day of week (mon, tue, wed). And I need to convert it in to a date.
How would I go about doing that in php?
Try this :
$yr="2014";
$week="01";
$day = "Mon";
echo date("Y-m-d", strtotime($yr."W".$week."D ".$day));
PHP fiddle:http://phpfiddle.org/main/code/nzg-ws6
Did a little digging and it looks like the day will need to be in numerical format for the strtotime conversion to work properly. Using a string such as 'Monday' won't work.
$yr = "2014";
$week = "01";
$day = "1"; //Monday
echo date("Y-m-d", strtotime($yr . "W" . $week . $day)); // "2014W011"
Note that I didn't include the 'D' prefix for the day as this looks to be following ISO8601.
Swapping your days to numbers can be done a few ways:
You could make an array Monday to Sunday and use the array key, or maybe a little easier you could do something like this:
$day date("N", strtotime("Monday")); // will echo 1
So the full code could look something like this:
$yr = "2014";
$week = "01";
$day = date("N", strtotime("Monday")); //
echo date("Y-m-d", strtotime($yr . "W" . $week . $day)); // "2014W011"
Here is a demo to check: http://3v4l.org/EllgQ
This converts you $day value to desired day format.
try this
$yr="2011";
$week="01";
$day="Monday";
$day=date("D",strtotime($day));
echo date("Y-m-d", strtotime($yr."W".$week."D ".$day));;
Demo
I have this piece of code which gets me a field from the database:
$end_date=$row1['end_date'];
If i print it it gives me something like: 25-09-2012
What i need is to get the month value, the year and date.
something like:
$month=09;
$day=25;
$year=2012;
How can i do that?
thanks!
Using DateTime:
$date = new DateTime($row1['end_date']);
$year = $date -> format('Y');
$month = $date -> format('m');
$day = $date -> format('d');
If your timestamps are all like the one provided, keep it simple:
list($day, $month, $year) = explode('-', $row1['end_date']);
In your case, you can use the explode function like this :
// store a string containing "25-09-2012"
$end_date = $row1['end_date'];
// split "25-09-2012" into an array of three elements
$thedate = explode("-", $end_date);
// retrieve the values
$month = $thedate[0]; // 25
$day = $thedate[1]; // 09
$year = $thedate[2]; // 2012
try
[month('end_date')]
[day('end_date')]
[year('end_date')]
Or use explode and use - as the delimiter
Take a peak at this helpful tutorial describing various formatting methods and useful date functions in PHP:
Date/Time Functions
Date Formats
A. You can use DateTime
$date = DateTime::createFromFormat('d-m-Y',$row1['end_date']);
$month = $date->format("m");
$day = $date->format("d");
$year = $date->format("Y");
B. Using strtotime
$date = strtotime($row1['end_date']);
$month = date("m", $date);
$day = date("d", $date);
$year = date("Y", $date);
C. You can just sscanf scan through the string
$date = sscanf($row1['end_date'], "%d-%d-%d");
$month = $date[0] ;
$day = $date[1] ;
$year = $date[2] ;
D. Another method is using list & explode
list($day, $month, $year) = explode('-', $row1['end_date']);
Do it on a single line and format it however you would like. (Dec, December, 12) and so on with date().
list($month, $day, $year) = explode('-', date('m-d-Y', strtotime($row1['end_date'])));
$values = getdate(strtotime($row1['end_date']));
echo $values['mon']; //month
echo $values['mday']; //day
echo $values['year']; //year
I usually mess this up, but I am trying to get a year from a variable $date = 2011-01-01 I am trying to get 2011 from the variable...I tried this but it didnt echo anything out...
date(strtotime("Y"), strtotime($date));
$year = date('Y', strtotime('2011-01-01'));
Or, if you don't want to convert to date, and trust your input format:
$year = explode('-', '2011-01-01')[0];
The above requires PHP 5.4, on older versions you need two lines:
$arr = explode('-', '2011-01-01');
$year = $arr[0];
Or yet, this slightly faster alternative suggested by tigrang:
$year = substr($date, 0, 4);
Or, an even faster way, as salathe suggested:
$year = (int) $date;
(everything after the first dash will be ignored, so $year will contain an integer with the year part of the date (unlike the other options, where $year would be a string).
date('Y', strtotime($date)) should do
First arg is the format, second is timestamp.
You can explode the date into a list and call each date item individually.
$date = '2012-01-01';
list($year, $month, $day) = explode('-', $date);
echo $year; // echoes 2012
May I suggest using PHP's DateTime class as an alternative:
$date = '2011-01-01';
$dateTime = new DateTime($date);
echo $dateTime->format('Y');
DateTime is all well and good, but lets go old school.
sscanf($date, '%d', $year);
// or
$year = strtok($date, '-');
// or
$year = substr($date, 0, 4);
// or
$year = (int) $date;
or, putting the sensible hat back on, you could use DateTime or correctly use date().
$datetime = new DateTime($date);
$year = $datetime->format('Y');
// or
$year = date('Y', strtotime($date));
I have a date in this format 030512 (ddmmyy).
But I'm having trouble with converting this to a date usable format I can add days to.
Basically.. I extracted the date above from a text file, Now I need to be able to add a number of days to it. But I am having trouble parsing the date in this format.
Is there another way of doing this rather then something like this:
// I have a date in this format
$date = '030512'; // 03 May 2012
$day = substr($date,0,2);
$month = substr($date, 2,2);
$year = substr($date, 4,2);
$date_after = $day . "-" . $month . "-".$year;
// Now i need to add x days to this
$total_no_nights = 010; // must be this format
$days_to_add = ltrim($total_no_nights,"0"); // 10, remove leading zero
// how do i add 10 days to this date.
You can do this (php >= 5.3):
$date = DateTime::createFromFormat('dmy', '030512');
$date->modify('+1 day');
echo $date->format('Y-m-d');
http://www.php.net/manual/en/datetime.createfromformat.php
For php < 5.3 :
$dateArray = str_split('030512', 2);
$dateArray[2] += 2000;
echo date("d/m/Y", strtotime('+1 day', strtotime(implode('-', array_reverse($dateArray)))));
try this using the month/day/year you already have:
$date = "$month/$day/$year";
$change = '+10 day';
echo date("d/m/Y", strtotime($change, strtotime($date)));
Assuming the date will always be in the future (or at least after 1st Jan 2000), you're not far wrong:
// I have a date in this format
$date = '030512'; // 03 May 2012
$day = substr($date,0,2);
$month = substr($date, 2,2);
$year = substr($date, 4,2);
// dd-mm-yy is not a universal format but we can use mktime which also gives us a timestamp to use for manipulation
$date_after = mktime( 0, 0, 0, $month, $day, $year );
// Now i need to add x days to this
$total_no_nights = "010"; // must be this format
$days_to_add = intval( $total_no_nights ); // No need to use ltrim
// Here's the "magic". Again it returns a timestamp
$new_date = strtotime( "+$days_to_add days", $date_after );
Using the DateTime object would be easier but you say you're not on PHP5.3.
You can't do date manipulation with strings becase, well, they are not dates. In PHP, you can use Unix timestamps (which are actually integers...) or DateTime objects. In you case:
$timestamp = strtotime('-10 days', mktime(0, 0, 0, $month, $day, $year));
echo date('r', $timestamp);
... or:
$object = new DateTime("$year-$month-$day");
$object->modify('-10 days');
echo $object->format('r');
Also, please note that 010 is an octal number that corresponds to 8 in decimal.
using the convert function in sql the date can be obtained in appropriate format.
anter that operations can be performed in php to manipulate the date. i hope that answers your query.
These answers don't quite do what I want:
Getting last month's date in php
How to find the last day of the month from date?
I have the following code that is meant to print a select-box with all the last day of month dates from a given date:
$last_case_date = $this->query_end_date();
$increm_date = "2011-06-01";
echo '<select name = "end_interval">';
$increm_date = date('Y-m-d', strtotime("$increm_date"));
while($increm_date <= $last_case_date) {
// next date is $increm_date + 1 month
$increm_date = date('Y-m-d', strtotime("$increm_date + 1 months"));
// then we want the last day of this new date
$month = substr($increm_date, 5, 2);
$year = substr($increm_date, 0, 4);
$increm_date = $this->last_day_of_month($month, $year);
echo '<option value = "'.$increm_date.'" selected = "selected"'.'>'.$increm_date.'</option>';
}
echo '</select>';
where last_day_of_month is this:
function last_day_of_month($month = '', $year = '') {
if(empty($month)) {
$month = date('m');
}
if(empty($year)) {
$year = date('Y');
}
$result = strtotime("{$year}-{$month}-01");
$result = strtotime('-1 second', strtotime('+1 months', $result));
return date('Y-m-d', $result);
}
which I borrowed from here.
Strange thing that happens is that I get these dates:
2011-07-31
2011-08-31
2011-10-31
but no 2011-09-30 ! It probably has something to do with September only having 30 days or something right?
Can anyone spot the problem? I've been staring at it for ages.... Thank you :).
You could just use the built in function for php
date('t', $timestamp);
So your function would look like:
function last_day_of_month($month = '', $year = '') {
$month
|| $month = date('m');
$year
|| $year = date('y');
$timestamp = mktime(0, 0, 0, $month, 1, $year);
return date('Y-m-t', $timestamp);
}
To get the last day of the month, use:
date('Y-m-t', strtotime("$year-$month-01"))
't' stands for the "number of days in the given month", i.e. for the last day. No need to add and subtract anything.
The below will output a select box with options for the last day of each month in the current year:
echo "<select>";
for($i=1;$i<=12;$i++){
echo "<option>".date('Y-m-t', strtotime(date("Y")."-".$i."-01"))."</option>";
}
echo "</select>";
If you want to set to a different year, simply replace date("Y") with, say, "2010". You could, for example loop through a number of years, running the code to output the options for each.
I think this is the problem of your piece of code:
the original $increm_date is "2011-06-01";
a a certain point in the while cycle $increm_date is 2011-08-31 (last day of august)
the in the cycle you add a month so $increm_date is 2011-10-01 (first day of october, NOT THE LAST OF SEPTMEBER) so you call the last_day_of_month providing OCTOBER NOT SEPTEMBER.
I am not sure of this. To test try to print the $increm_date before last_day_of_month is called