I am getting ISO datetime using below code
$date= date("c");
echo $date;
which returns something like below
2016-01-07T20:18:46+00:00
But the API I use tells that its wrong format and it needs in below format:
2016-01-07T20:35:06+00Z
I need to remove :00 at the end and add Z.
I am completely new to regex , can anyone help me understand the regex and tell which format is required.
You'll want to define the date format specifically.
If microseconds will always be 00
date("Y-m-d H:i:s+00\Z");
Else, use this little big of logic
date("Y-m-d H:i:s+"). substr((string)microtime(), 2, 2) . 'Z';
More info.
http://php.net/manual/en/function.date.php
You can find the last occurrence of : with strrpos, and get the substring up to it with substr, and then add Z:
$date= date("c");
echo "Current: $date\n"; // => 2016-01-07T21:58:08+00:00
$new_date = substr($date, 0, strrpos($date, ":")) . "Z";
echo "New : " . $new_date; // => 2016-01-07T22:10:54+00Z
See demo
Related
I am dealing with a problem of time conversion from 12 hr format to 24 hour format.
Is there any single function in php to replace the first two characters of a string?
str_replace can be used only when I know the substring content to be replaced.
$str_to_replace = '12';
$input_str = 'ab345678';
$output_str = $str_to_replace . substr($input_str, 2);
echo $output_str;
"12345678"
If the date is always given in a specific format you could try to convert it to a DateTime object and format the output.
$dateString = '15-Feb-2009 2:24 PM';
$date = DateTime::createFromFormat('j-M-Y g:i A', $dateString);
echo $date->format('Y-m-d G:i'); // will show "2009-02-15 14:24"
In general you should try in avoid holding a date in a string. Convert it to a DateTime -- this makes it also easier for you to manipulate the object (e.g. move date +1 day)
My intention is to convert the following date
20/04/17 13:27:5
to this
20-04-2017 13:27:05
I tried the typical date format functions of php and also Carbon...
things like
$newDate= Carbon::createFromFormat('d/m/y H:m:s', $originalDate);
in this case
var_dump($newDate->toDateTimeString()) would bring 2019-03-20 13:00:55 which is not what I expect.
So I was not lucky....is there a way to do this in a straight forward manner?
I think this should work.
$date = "20/04/17 13:27:5";
$sec = substr($date, strrpos($date, ":") + 1);
$sec = substr("0{$sec}", -2);
$new = substr($date, 0, strrpos($date, ":") + 1) . $sec;
$newDate = Carbon::createFromFormat('d/m/y H:i:s', $new);
I changed the format since you were using m twice for "minutes" and "month". It is correct for the month, but not for the minutes. Instead use i for minutes with leading zeroes.
$sec Is what I used to get the second from the string. This gets the last position of : and will take everything after it. This assumes that you do not change the format of the string.
substr("0{$sec}", -2) Adds a zero to the current second and extracts the last two characters. That means that 50 becomes 050 and then the last two characters are 50 so we end up without the padding, but 5 becomes 05 and the last two characters are the only characters.
$new concatenates the start of the date string and the new second with the zero padding.
$newDate is your original string with the format changed.
There is issue with seconds. There must be 05 not only 5
<?php
$original_date = "20/04/17 13:27:5";
$date_explode = explode(":", $original_date);
$date_explode[2] = str_pad($date_explode[2],2,"0",STR_PAD_LEFT);
$original_date = implode($date_explode,":");
$date = DateTime::createFromFormat('d/m/y H:i:s', $original_date);
echo date_format($date,"d-m-Y H:i:s");
?>
This is a working conversion routine that creates the ISO format you are looking for. But as already mentioned you need to "fix" the strange way the seconds are specified in the original example you provide. You will have to use string functions if that really is the format you receive. Better would be to fix the code that creates such broken formats.
<?php
$input = '20/04/17 13:27:05';
$date = DateTime::createFromFormat('d/m/y H:i:s', $input);
var_dump($date->format('d-m-Y H:i:s'));
The output obviously is:
string(19) "20-04-2017 13:27:05"
Isn't it like this?
$newDate = Carbon::createFromFormat('d/m/y H:i:s', $originalDate);
This question already has answers here:
Parse and reformat a datetime string
(6 answers)
Closed 11 months ago.
I have a string with a date which is in this format MMDDYYYY (ie. 01132012, 01142012 etc.)
I need to do something on a page, if that string is 14 days or less from the current date.
ie. Today is 01132012, so any strings with 12312011 or a less date are going to be showing something on a page.
Can anyone help with this? I've tried
echo date("d/m/Y", strtotime('01142012'));
But to no avail.
You can use the DateTime class of PHP
<?
// current date
$now = new DateTime();
//your date
$date = DateTime::createFromFormat('mdY', '01142012');
// calculate difference
$diff = $now->diff($date);
...
// output the date in format you want
echo $date->format('d/m/Y');
?>
EDIT: I just realized, that your format isn't one supported by php. So you have to use alternate objectbuild.
I prefer using strptime.
<?
$dt = strptime('01142012', '%m%d%Y');
echo sprintf("%02d/%02d/%04d", $dt['tm_mday'], $dt['tm_mon']+1, $dt['tm_year']+1900);
If you use PHP 5.3 or above, you can also use date_parse_from_format()
How about some substr + mktime?
$string = '01142012';
$time = mktime(0, 0, 0,
substr($string, 0, 2),
substr($string, 2, 2),
substr($string, 4, 4)
);
echo date('d/m/Y', $time);
try date('m-d-y', strtotime('01142012'));
could also try something like;
$var = strtotime('01142012');
$var2 = date ('F j, Y', $var);
Your string input of '01142012' cannot be parsed by strtotime() as it is not a valid as it is returning -1 as an answer. To convert this into a valid date you will need to add either slashes or dashes to separate the numbers.
The easiest way would be to store the dates with the dashes or slashes, such as '01-14-2012' or '01/14/2012' in the database from now on or you are going to have to create your own function to convert the numbers into a valid form for strtotime().
To do this you could do something like this:
function makeValidDate($date) {
$valid_date = array();
$array = str_split($date); //split characters up
foreach($array as $key => $character){
if($key==2 || $key==4){
$character = '-'.$character; //add relevant formatting to date
$valid_date[] = $character; //add this to the formatted array
}
else{
$valid_date[] = $character; // if not dashes or slashes needed add to valid array
}
}
return implode($valid_date); // return the formmatted date for use with strtotime
}
You can then do this to get a valid date:
$valid_date = makeValidDate('01142012');
echo date("d/m/Y", strtotime($valid_date));
I haven't tested this but you should get a good idea of what to do.
EDIT: Capi's idea is a lot cleaner!!
try "preg_match(pattern,string on wich the pattern will be aplied)";
http://www.php.net/manual/en/function.preg-match.php
you can also define an offset. so first take te first 2 digits. than take the other 2 digits and after that get the other four digits. after that place them in one string. after that use maketime,strtotime,date. this kind of stupid solution but i only thought of that. hope this will help
I am building a timestamp from the date, month and year values entered by users.
Suppose that the user inputs some wrong values and the date is "31-02-2012" which does not exist, then I have to get a false return. But here its converting it to another date nearby. Precisely to: "02-03-2012"..
I dont want this to happen..
$str = "31-02-2012";
echo date("d-m-Y",strtotime($str)); // Outputs 02-03-2012
Can anyone help? I dont want a timestamp to be returned if the date is not original.
You might look into checkdate.
That's because strtotime() has troubles with - since they are used to denote phrase like -1 week, etc...
Try
$str = '31-02-2012';
echo date('d-m-Y', strtotime(str_replace('-', '/', $str)));
However 31-02-2012 is not a valid English format, it should be 02-31-2012.
If you have PHP >= 5.3, you can use createFromFormat:
$str = '31-02-2012';
$d = DateTime::createFromFormat('d-m-Y', $str);
echo $d->format('d-m-Y');
You'll have to check if the date is possible before using strtotime. Strtotime will convert it to unix date meaning it will use seconds since... This means it will always be a date.
You can workaround this behavior
<?php
$str = "31-02-2012";
$unix = strtotime($str);
echo date('d-m-Y', $unix);
if (date('d-m-Y', $unix) != $str){
echo "wrong";
}
else{
echo date("d-m-Y", $unx);
}
or just use checkdate()
Use the checkdate function.
$str = "31-02-2012";
$years = explode("-", $str);
$valid_date = checkdate($years[1], $years[0], $years[2]);
Checkdate Function - PHP Manual & Explode Function - PHP Manual
Combine date_parse and checkdate to check if it's a valid time.
<?php
date_default_timezone_set('America/Chicago');
function is_valid_date($str) {
$date = date_parse($str);
return checkdate($date['month'], $date['day'], $date['year']);
}
print is_valid_date('31-02-2012') ? 'Yes' : 'No';
print "\n";
print is_valid_date('28-02-2012') ? 'Yes' : 'No';
print "\n";
Even though that date format is acceptable according to PHP date formats, it may still cause issues for date parsers because it's easy to confuse the month and day. For example, 02-03-2012, it's hard to tell if 02 is the month or the day. It's better to use the other more specific date parser examples here to first parse the date then check it with checkdate.
I'm working on a project that requires me to read values from a file and manipulate them slightly before storing them elsewhere. One of the things I need to do is convert some dates from mm/dd/yy format to mm/dd/yyyy format. Unfortunately for me, I am relatively new to PHP and regular expressions (which I assume is one of the better ways to solve this problem), and am therefore somewhat mystified. Any and all help will be appreciated. Thanks!
PHP has a built-in function strtotime() that is meant for just this kind of task... it'll even do the best-guess for the 2-digit year following this rule: the values between 00-69 are mapped to 2000-2069 and 70-99 to 1970-1999. Once you have the date/time in the UNIXy format that PHP prefers, then you can print it out however you want with the date() function.
<?php
$str = '02/28/98';
// in PHP 5.1.0+, strtotime() returns false if it fails
// previous to PHP 5.1.0, you would compare with -1 instead of false
if (($timestamp = strtotime($str)) === false) {
echo "Couldn't figure out the date ($str)!";
} else {
echo "Reformatted date is " . date('m/d/Y', $timestamp);
}
?>
(I presume we're timezone-agnostic here, or that would add complications.)
You can try this, it may or may not work:
$new_date = date( 'm/d/Y', strtotime( $old_date ) );
Where $old_date is in the format you're talking about.
One of the problems here is that YY, assuming it is relatively current, can be either 19YY or 20YY. This means you should pick a number to be the cut off. Let's call this $cutOff If YY is less than $cutOff, we want 20YY if greater than we want 19YY
You could do it with regex, but since your example is simple and regex tends to be slower, you can simply use string manipulation with substr and substr_replace.
Here's how to change a string mm/dd/yy int mm/dd/yyyy:
<?php
// Our date
$str = "01/04/10";
$cutoff = 50;
// See what YY is
// Get the substring of $str starting two from the end (-2)... this is YY
$year = substr($str, -2);
// Check whether year added should be 19 or 20
if ($year < 50)
// PHP converts string to number nicely, so this is our 20YY
$year += 2000;
else
// This is 19YY
$year += 1900;
// Repace YY with YYYY
// This will take $str and replace the part two from the end (-2) undtil
// the end with $year.
$str = substr_replace($str, $year, -2);
// See what we got
echo $str;
?>
You can append the year starting two values (19 or 20) as below:
//for $s_date = "yy-dd-mm"
if (substr($s_date,6,2) >= 50){
$standarddate = "19" . substr($s_date,6,2); //19yy
$standarddate .= "-" . substr($s_date,0,2); //mm
$standarddate .= "-" . substr($s_date,3,2); //dd
} else {
$standarddate = "20" . substr($s_date,6,2); //20yy
$standarddate .= "-" . substr($s_date,0,2); //mm
$standarddate .= "-" . substr($s_date,3,2); //dd
}
// you will get yyyy-mm-dd