I have a problem in this code:
$month = $_POST['month'];
$day = $_POST['day'];
$year = $_POST['year'];
function dateImplodeFunction($year, $month, $day){
$array = array($year, $month, $day);
$date = date('Y-m-d', strtotime( implode("-", $array)));
return $date;
}
based on the code above I'm going to create a function where there are 3 inputs month, day and a year. When I input those 3 these variables will passed to this function, combine those 3 variables and use the implode function to create a format based on what date you specified. for instance let's say if I input 10/01/1989 it will echo the display 10/01/1989.
also I need to use date function together with strtotime function (refer to the code above) for database, setting my date field into date data type.
The problem here is if I input 10-01-1989, it returns/dipslays the value of 01-01-1970 why??
I figured it out that there is a conflict between strtotime and implode function due to test. I've search through google but it find none. I hope you can help me. Thanks in advance.
sorry for the bad english =P
I would just use the Date Time class.
$month = '07';
$day = '26';
$year = '2012';
$timezone = new DateTimeZone('America/New_York');
$date = new DateTime( "{$year}-{$month}-{$day}", $timezone );
print $date->format('Y-m-d H:i:s');
# Output: 2012-07-26 00:00:00
http://codepad.org/VxwDHPeU
You should use mktime :
function dateImplodeFunction($year, $month, $day)
{
return date('Y-m-d', mktime(0, 0, 0, $month, $day, $year));
}
Documentation of mktime here
Related
I am trying to display the date that I send by $_GET, my url format is for example http://127.0.0.1/index.php?date=01/08/2018.
The result is 08-01-2018 it change the order of the day and the month why ?
$orginal_date= $_GET['date'];
$date = date("d-m-Y", strtotime($original_date));
echo $date;
It automatically parses the date with format m/d/Y (american format). You could use the DateTime class to specify your format:
$date = DateTime::createFromFormat('d/m/Y', $_GET['date']);
echo $date->format('d-m-Y');
Documentation:
DateTime::createFromFormat
DateTime::format
list($month, $day, $year) = explode("/", $_GET["date"]);
$date = sprintf("%d-%02d-%02d", $year, $month, $day);
This looks like you may be assuming how php will use strtotime and convert to a timestamp. By default it thinks that 01/08/2018 is Jan 8 2018. There is a method date_create_from_format that you can tell it the format you are expecting and then you can format that to however you want it to look like.
$orig = $_GET['date'];
$date = date_create_from_format('d/m/Y', $orig);
echo date_format($date, 'd-m-Y');
//result is 01-08-2018
Try This
list($month, $day, $year) = explode("/", $_GET["date"]);
echo date('d-m-Y',mktime(0, 0, 0, $month,$day, $year));
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 string with following format:
$date = "2012-07-22 17:48:24";
I want to get the year, month and date in the variables and ignore the time. I am trying following:
list($year, $month, $day) = split('[-]', $date);
This returns correct values to $year and $month, but the $day gets: "22 17:48:24", while I want to get only 22.
Instead of exploding the value you could use a DateTime object:
<?php
$date = "2012-07-22 17:48:24";
$dateTime = new DateTime($date);
var_dump(array(
'year' => $dateTime->format('Y'),
'month' => $dateTime->format('m'),
'day' => $dateTime->format('d'),
));
This would be the most flexible option imho.
As #zerkms noted in his comment you could also use strtotime() and date(), but I find myself only using the DateTime class lately. Not only because it has a nice OO API, but also because it will keep on working after the year 2038 :-). The comment is not wrong though.
There's also the sscanf() function.
sscanf('2012-07-22 17:48:24', '%d-%d-%d', $year, $month, $day);
Use explode:
$date = "2012-07-22 17:48:24";
$date = explode(" ", $date);
list($year, $month, $day) = split('[-]', $date[0]);
EDIT:
You should use explode for the date too:
list($year, $month, $day) = explode('-', $date[0]);
The use of split is discouraged as it was deprecated.
$date = "2012-07-22 17:48:24";
preg_match('~^(\d{4})-(\d{2})-(\d{2}) (\d{2}):(\d{2}):(\d{2})$~', $date, $m);
print_r($m);
$date = date('Y-m-d', strtoime("2012-07-22 17:48:24"));
list($year, $month, $day) = split('[-]', $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.