$doba = explode("/", $dob);
$date = date("Y-m-d", mktime(0,0,0, $doba[0], $doba[1], $doba[2]));
The above code turns any date i pass through into 1999-11-30 and i know it was working yesterday. Date is correct when I echo $doba. Anyone have any ideas?
Cheers
What is the format of $doba? Remember mktime's syntax goes hour, minute, second, month, day year which can be confusing.
Here's some examples:
$doba = explode('/', '1991/08/03');
echo(date('Y-m-d', mktime(0,0,0, $doba[1], $doba[2], $doba[0]);
$doba = explode('/', '03/08/1991');
echo(date('Y-m-d', mktime(0,0,0, $doba[1], $doba[0], $doba[2]);
or even easier: $date = date('Y-m-d', strtotime($dob))
It is a bit overkill to use mktime in this case. Assuming $dob is in the following format:
MM/DD/YYYY
you could just to the following to acheive the same result (assuming $dob is always valid):
$doba = explode("/", $dob);
$date = vsprintf('%3$04d-%1$02d-%2$02d', $doba);
If you have issues with what jcoby said above, the strptime() command gives you more control by allowing you to specify the format as well.
Related
I'm upload the excel data into mysql. in there date save like 16.5.59(dd.mm.yy) formate. I try to change this using this code
$date = '16.5.59';
echo date('d-m-y',strtotime($date));
it always show current time like 04-02-15.
Pls change this to dd-mm-yyyy formate.
Thanks in Advance.
Timestamp was limited from 01-01-1970 to 19-01-2038 on some systems (e.g. Windows).
If it not cover the between given range then it will take current date.
$date = '16.5.59';
$date_array = explode(".",$date);
$var_day = $date_array[0];
$var_month = $date_array[1];
> Blockquote
$var_year = $date_array[2];
echo $new_date_format = "$var_day-$var_month-$var_year";
Try to make the date yourself using explode and strtotime:
date_default_timezone_set('America/Los_Angeles');
$date = '16.5.59';
$date_time = strtotime(implode('-', array_reverse(explode('.', $date))))
$date_str = date('d-m-Y', $date_time);
echo $date_str;
Output: 16-5-2059
You ARE required to specify that the year is 1959 instead of 2059 (or are you sure that it's really 2059?)
You can make change from 2059 to 1959 like this:
date_default_timezone_set('America/Los_Angeles');
$date = '16.5.59';
$date_arr = array_reverse(explode('.', $date))
# This line does the job. Make sure all years are between 1900 - 1999.
$date_arr[0] = '19' . $date_arr[0];
$date_time = strtotime(implode('-', $date_arr))
$date_str = date('d-m-Y', $date_time);
echo $date_str;
I hope this answer can help you.
This is by no means the best solution but using php's date_parse_from_format is possible. You could supply a format and use the following
$date = "16.5.59";
$dateObj = date_parse_from_format("j.n.yy", $date)
with the $dateObj you can work what you need such as :
echo $dateObj['year'];
IDE Running Example
Worth noting you require PHP >=v5.3
You can do this if it is for simpler task. It will consume time.
$date = '16.5.59';
$dtSplit=explode(".",$date);
echo "<br>".$dtSplit[0].".".$dtSplit[1].".".$dtSplit[2];
echo "<br>".$dtSplit[0].".".$dtSplit[2].".".$dtSplit[1];
echo "<br>".$dtSplit[1].".".$dtSplit[0].".".$dtSplit[2];
echo "<br>".$dtSplit[1].".".$dtSplit[2].".".$dtSplit[0];
echo "<br>".$dtSplit[2].".".$dtSplit[0].".".$dtSplit[1];
echo "<br>".$dtSplit[2].".".$dtSplit[1].".".$dtSplit[0];
I have the following value in a variable: 26/03/2011
I'd like to get this in the format: 2011-03-26
How do I achieve this?
Many thanks
Try this:
list($d, $m, $y) = explode("/", "26/03/2011");
echo "$y-$m-$d";
While there are many ways to do this, I think the easiest to understand and apply to all date conversions is:
$date = date_create_from_format('d/n/Y', $date)->format('Y-n-d');
It is explicit and you'll never have to wonder about m/d or d/m, etc.
Might be a good idea to use the date() function combined with mktime():
$date = explode('/', '26/03/2011');
echo date('Y-m-d', mktime(0,0,0,$date[1],$date[0],$date[2]));
The reason why this could be a good idea is if you ever want to format the date in a different (more complex) format, say "March 26th, 2011" you could just do:
$date = explode('/', '26/03/2011');
echo date('F jS, Y', mktime(0,0,0,$date[1],$date[0],$date[2]));
I'm having date 20/12/2001 in this formate . i need to convert in following format 2001/12/20 using php .
$var = explode('/',$date);
$var = array_reverse($var);
$final = implode('/',$var);
Your safest bet
<?php
$input = '20/12/2001';
list($day, $month, $year) = explode('/',$input);
$output= "$year/$month/$day";
echo $output."\n";
Add validation as needed/desired. You input date isn't a known valid date format, so strToTime won't work.
Alternately, you could use mktime to create a date once you had the day, month, and year, and then use date to format it.
If you're getting the date string from somewhere else (as opposed to generating it yourself) and need to reformat it:
$date = '20/12/2001';
preg_replace('!(\d+)/(\d+)/(\d+)!', '$3/$2/$1', $date);
If you need the date for other purposes and are running PHP >= 5.3.0:
$when = DateTime::createFromFormat('d/m/Y', $date);
$when->format('Y/m/d');
// $when can be used for all sorts of things
You will need to manually parse it.
Split/explode text on "/".
Check you have three elements.
Do other basic checks that you have day in [0], month in [1] and year in [2] (that mostly means checking they're numbers and int he correct range)
Put them together again.
$today = date("Y/m/d");
I believe that should work... Someone correct me if I am wrong.
You can use sscanf in order to parse and reorder the parts of the date:
$theDate = '20/12/2001';
$newDate = join(sscanf($theDate, '%3$2s/%2$2s/%1$4s'), '/');
assert($newDate == '2001/12/20');
Or, if you are using PHP 5.3, you can use the DateTime object to do the converting:
$theDate = '20/12/2001';
$date = DateTime::createFromFormat('d/m/Y', $theDate);
$newDate = $date->format('Y/m/d');
assert($newDate == '2001/12/20');
$date = Date::CreateFromFormat('20/12/2001', 'd/m/Y');
$newdate = $date->format('Y/m/d');
I want split a string using PHP to get a individual date info.
For example:
$date = '08/05/2010';
would create the varables
$month = '08';
$day = '05';
$year = '2010';
Thank you.
Use explode:
$date = '08/05/2010';
list($month, $day, $year) = explode('/', $date);
if that's your example, you could explode it into an array.
$array = explode('/', $date);
list($month, $day, $year) = explode('/', $date);
Assuming it's always in that format, you want explode:
<?php
$date = '08/05/2010';
$arr = explode("/", $date);
list($month, $day, $year) = $arr;
// $month = 08, $day = 05, $year = 2010.
?>
The answers posted above would do the trick. You probably also want to check the date conforms to your expected format before you run your function. The checkdate function would be useful for this, or this snippet is a standalone implementation.
I agree with the comments about using the checkdate function, however that is really for after you split the date apart because you pass in each part (month, day, year) to check.
If you are getting the original date from user input you might want to make sure that you are getting the right format before you separate it out (using list or explode as shown previously.)
I see two options for validity if it is coming from a user: first is to only allow them to select the date, and not freely enter it. The second is to make sure you are getting it in the mm/dd/yyyy format you are expecting. You could perform a regular expression match on it before the separation. The regular expression could be something like /\d{2}/\d{2}/\d{4}/
Anyway, not really needed if you are sure of your source, but it is always important to think of where your data is coming from and how clean it is.
How can I convert this string 05/Feb/2010:14:00:01 to unixtime ?
Use the strtotime function:
Example:
$date = "25 december 2009";
$my_date = date('m/d/y', strtotime($date));
echo $my_date;
For PHP 5.3 this should work. You may need to fiddle with passing $dateInfo['is_dst'], wasn't working for me anyhow.
$date = '05/Feb/2010:14:00:01';
$dateInfo = date_parse_from_format('d/M/Y:H:i:s', $date);
$unixTimestamp = mktime(
$dateInfo['hour'], $dateInfo['minute'], $dateInfo['second'],
$dateInfo['month'], $dateInfo['day'], $dateInfo['year'],
$dateInfo['is_dst']
);
Versions prior, this should work.
$date = '05/Feb/2010:14:00:01';
$format = '#^(?P<day>\d{2})/(?P<month>[A-Z][a-z]{2})/(?P<year>\d{4}):(?P<hour>\d{2}):(?P<minute>\d{2}):(?P<second>\d{2})$#';
preg_match($format, $date, $dateInfo);
$unixTimestamp = mktime(
$dateInfo['hour'], $dateInfo['minute'], $dateInfo['second'],
date('n', strtotime($dateInfo['month'])), $dateInfo['day'], $dateInfo['year'],
date('I')
);
You may not like regular expressions. You could annotate it, of course, but not everyone likes that either. So, this is an alternative.
$day = $date[0].$date[1];
$month = date('n', strtotime($date[3].$date[4].$date[5]));
$year = $date[7].$date[8].$date[9].$date[10];
$hour = $date[12].$date[13];
$minute = $date[15].$date[16];
$second = $date[18].$date[19];
Or substr, or explode, whatever you wish to parse that string.
You should look into the strtotime() function.
http://www.php.net/date_parse_from_format
$d="05/Feb/2010:14:00:01";
$dr= date_create_from_format('d/M/Y:H:i:s', $d);
echo $dr->format('Y-m-d H:i:s');
here you get date string, give format specifier in ->format() according to format needed
Simple exploding should do the trick:
$monthNamesToInt = array('Jan'=>1,'Feb'=>2, 'Mar'=>3 /*, [...]*/ );
$datetime = '05/Feb/2010:14:00:01';
list($date,$hour,$minute,$second) = explode(':',$datetime);
list($day,$month,$year) = explode('/',$date);
$unixtime = mktime((int)$hour, (int)$minute, (int)$second, $monthNamesToInt[$month], (int)$day, (int)$year);
If you're up for it, use the DateTime class
Try this:
$new_date=date('d-m-Y', strtotime($date));
If it's a string that you trust meaning that you have checked it before hand then the following would also work.
$date = new DateTime('2015-03-27');