I want to parse the date 1938 1938+02:00 using date() & strtotime().
My code:
echo date("Y", strtotime("1938+02:00"));
gives me as result "2014"..
What am i doing wrong?
For something like this just get the first four characters of the string. No need to work with dates and such:
echo substr('1938+02:00', 0, 4);
Demo
But if you insist on using date functionality you'll need to use DateTime::createFromFormat() as that date string is not a standard format.
$date = DateTime::createFromFormat('YP', '1938+02:00');
echo $date->format('Y');
Demo
date("Y"); only return the year. That's what the Y does.
See the Manual page for date for other options.
EDIT
Another thing to consider, is that timestamps only go back to 1970. That's what a timestamp is (the number of seconds since 1970).
So, that's going to give you a negative value for the timestamp.
Your date string is not in an acceptable format. here is a list of acceptable formats for strtotime
Related
I have a php string from db it is 20/11/2017 I want to convert it milliseconds.
It's my code to doing that.
$the_date = "20/11/2017";
$mill_sec_date = strtotime($the_date);
var_dump($mill_sec_date);
But it does not print any thing rather than
bool(false);
What is the problem and how can i solve it ????
When using slashes to separate parts of the date, PHP recognizes the format as MM/DD/YYYY. Which makes your date invalid because there is no 20th month. If you want to use the format where day and month is swapped, you need to use hyphens, like DD-MM-YYYY.
$time = strtotime('10/16/2003');
$newformat = date('Y-m-d',$time);
print_r($newformat);
Use DateTime class to call function createFromFormat
$date = date_create_from_format('d/M/Y:H:i:s', $string);
$date->getTimestamp();
Most likely you got the date format wrong, see
here for a list of supported date and time formats:
This section describes all the different formats that the strtotime(), DateTime and date_create() parser understands.
You string is not accept by the strtotime, you can use createFromFormat set set the with the format type of the time string like below, you can also check the live demo. And you also can refer to this answer
var_dump(DateTime::createFromFormat('d/m/Y', "20/11/2017"));
I am using a variable to get the date from the database,then i am passing that variable to strtotime function to get the desired format,but it always returning wrong date.perhaps there is a problem in passing a variable in strtotime function.please suggest me guys,how should i get the correct date in correct format.
Here is what i am trying to do
$date = $fetch_user['date'];
$newDate = date("d-m-Y", strtotime($date));
$day = date('l', strtotime($newDate));
echo $newDate;
echo "-----";
echo $day;
exit;
January 1, 1970 is the so called Unix epoch. It's the date where they started counting the Unix time. If you get this date as a return value, it usually means that the conversion of your date to the Unix timestamp returned a (near-) zero result. So the date conversion doesn't succeed. Most likely because it receives a wrong input.
In other words, your strtotime($date) returns 0, meaning that $date is passed in an unsupported format for the strtotime function.
So you'll have to check for yourself $date, before calling strtotime at all.
01-01-1970 means you probably get 0 as a result of strtotime(). You are probably using a format, that this function cannot understand. The PHP documentation states:
The function expects to be given a string containing an English date
format and will try to parse that format into a Unix timestamp (the
number of seconds since January 1 1970 00:00:00 UTC), relative to the
timestamp given in now, or the current time if now is not supplied.
So it is not really flexible. You might want to try DateTime::createFromFormat instead. Take a look at it's documentation.
Basically you have to specify the format of the date string, that you give as an input as well. That way you can use whatever date format you want.
Example from php.net:
<?php
$date = DateTime::createFromFormat('j-M-Y', '15-Feb-2009');
I want to convert date 24/09/2010 in format dd/mm/yyyy to 2010-09-24 in format yyyy-mm-dd.
This works:
date("Y-m-d",strtotime("09/24/2010"));
But this does not:
date("Y-m-d",strtotime("24/09/2010")); // it returns '1970-01-01'
Any idea why?
according to php, the valid php formats are stated here. So basically what you gave is invalid.
Alternatively, you can use mktime, date_parse_from_format or date_create_from_format
strtotime does its best to guess what you mean when given a string, but it can't handle all date formats. In you example, it is probably thinking that you are trying to refer to the 24th month, which isn't valid, and returns 0, which date then treats as the unix epoch (the date you got).
you can get around this using the mktime() and explode() functions, like so:
$date = "24/09/2010";
$dateArr = explode("/",$date);
$timeStamp = mktime(0,0,0,$dateArr[1],$dateArr[0],$dateArr[2]);
$newFormat = date("Y-m-d",$timeStamp);
As you say,
date("Y-m-d",strtotime("09/24/2010"))
will work,because the date format--"09/24/2010"is correct,
but "24/09/2010" is not the correct date format.
you can find something useful here
Hallo, I want to find the difference between today and another date,
convert todays date into unix time format here
<?php
echo '1st one'.strtotime(date("d.m.Y")).'<br>';
echo '2nd one'.strtotime(date("m.d.Y")).'<br>';
?>
The first echo is producing some value, but not the second one. What is the bug in it...please help..
strtotime makes assumptions based on the date format you give it. For instance
date("Y-m-d", strtotime(date("d.m.Y"))) //=> "2010-09-27"
date("Y-m-d", strtotime(date("m.d.Y"))) //=> "1969-12-31"
Note that when given an invalid date, strtotime defaults to the timestamp for 1969-12-31 19:00:00, so when you end up with an unexpected date in 1969, you know you're working with an invalid date.
Because strtotime is looking for day.month.year when you use . as the delimiter, so it sees "9.27.2010" as the 9th day of the 27th month, which obviously doesn't exist.
However, if you change it to use / as the delimiter:
date("Y-m-d", strtotime(date("d/m/Y"))) //=> "1969-12-31"
date("Y-m-d", strtotime(date("m/d/Y"))) //=> "2010-09-27"
In this case, strtotime expects dates in month/day/year format.
If you want to be safe, Y-m-d is generally a good format to use.
It's worth pointing out that strtotime() does accept words like "today" as valid input, so you don't need to put a call to date() in there if all you want is today's date. You could just use strtotime('today');.
Come to think of it, a simple call to time(); will get you the current time stamp too.
But to actually answer the question, you need to consider that d.m.Y and m.d.Y are ambiguous - if the day of the month is less than the 12th, it is impossible to tell which of those two date formats was intended. Therefore PHP only accepts one of them (I believe it uses m/d/Y if you have slashes, but for dots or dashes it assumes d-m-Y.
If you're using strtotime() internally for converting date formats, etc, there is almost certainly a better way to do it. But if you really need to do this, then use 'Y-m-d' format, because it's much more universally reliable.
On the other hand, if you're accepting date input from your users and assuming that strtotime() will deal with anything thrown at it, then sadly you're wrong; strtotime() has some quite big limitations, of which you've found one. But there are a number of others. If you plan to use strtotime() for this sort of thing then you need to do additional processing as well. There may also be better options such as using a front-end Javascript date control to make it easier for your users without having to rely on strtotime() to work out what they meant.
strtotime does not consider 09.27.2010 to be a valid date...
You could check it like this:
<?php
// will return false (as defined by the docs)
echo var_dump(strtotime("09.27.2010"));
?>
The function expects to be given a string containing a US English date format and will try to parse that format into a Unix timestamp. US time format is : MM DD YYYY
look here for the Information about which formats are valid http://www.php.net/manual/en/datetime.formats.php. But what do you mean with deference between 2 dates? You mean the Timespan between 2 dates?
echo (time() - strotime("- 2 days")) . " seconds difference";
Something like that?
strtotime would not take the d.m.y format. good way is Y-m-d
Duplicate
Managing date formats differences between PHP and MySQL
PHP/MySQL: Convert from YYYY-MM-DD to DD Month, YYYY?
Format DATETIME column using PHP after printing
date formatting in php
Dear All,
I have a PHP page where i wil be displaying some data from Mysql db.
I have 2 dates to display on this page.In my db table, Date 1 is in the format d/m/Y (ex: 11/11/2002) and Date 2 is in the format d-m-Y (ex : 11-11-2002)
I need to display both of this in the same format .The format i have stored in a variable $dateFormat='m/d/Y'
Can any one guide me
Thanks in advance
Use strtotime to convert the strings into a Unix timestamp, then use the date function to generate the correct output format.
Since you're using the UK date format "d/m/Y", and strtotime expects a US format, you need to convert it slighly differently:
$date1 = "28/04/2009";
$date2 = "28-04-2009";
function ukStrToTime($str) {
return strtotime(preg_replace("/^([0-9]{1,2})[\/\. -]+([0-9]{1,2})[\/\. -]+([0-9]{1,4})/", "\\2/\\1/\\3", $str));
}
$date1 = date($dateFormat, ukStrToTime($date1));
$date2 = date($dateFormat, ukStrToTime($date2));
You should be all set with this:
echo date($dateFormat, strtotime($date1));
echo date($dateFormat, strtotime($date2));
You may want to look into the strptime function. This can convert any date from a string back into numeric values. Unlike strtotime, it can be adapted to different formats, including those from different locales, and its output is not a UNIX timestamp, so it's capable of parsing dates before 1970 and after 2037. It may be a little bit more work though because it returns an associative array though.
Unfortunately it's not available on Windows systems either so it's not portable.
If for some reason strtotime will not work for you, could always just replace the offending punctuation with str_replace.
function dateFormat($date) {
$newDate = str_replace(/, -, $date);
echo $newDate;
}
echo dateFormat($date1);
echo dateFormat($date2);
I know this will make most folks cringe, but it may help you with formatting non-date strings in the future.
rookie i am. so came up with the method that just do that. what mysql needs.. shish i used param 2... hope it helps. regards
public function dateConvert($date,$param){
if($param==1){
list($day,$month,$year)=split('[/.-]',$date);
$date="$year-$month-$day"; //changed this line
return $date;
}
if ($param == 2){ //output conversion
list($day,$month,$year) = split('[/.]', $date);
$date = "$year-$day-$month";
return $date;
}
}