This should be simple, but I'm having trouble... so turn to StackOverflow...
I'm in the UK and am getting a date from the jQuery DatePicker in a dd/mm/yy format.
I want to store this date as a serial (yyyymmdd) so run a Date_To_Serial function that just does:
return date("Ymd", strtotime($strDate_In));
where $strDate_In is the date string in dd/mm/yy format.
So, passing in 01/12/2013, I expect 20131201 to be returned.
But, when 01/12/2013 is passed in, it returns 20130112 - PHP obviously assumes the date format is mm/dd/yyyy.
How can I fix this please?
If you know the format, try using the static createFromFormat method on the DateTime class:
$date = DateTime::createFromFormat('d/m/Y', '01/12/2013');
return $date->format('Ymd');
If the separator is a / then strtotime assumes a US-format. To have it recognise a UK format the separator must be - or .:
echo date('Ymd', strtotime('01/12/2014')); // 20140112
echo date('Ymd', strtotime('01-12-2014')); // 20141201
So for this to work in your example you would do:
return date("Ymd", strtotime(str_replace('/', '-', $strDate_In)));
Use a DateTime object's createFromFormat method. It allows you to specify the format of the input.
Afterwards, use the format method to export the date in the desired format.
Check out DateTime::createFromFormat for correct handling of non-standard date formats.
return DateTime::createFromFormat("d/m/Y", $strDate_In)->format("Ymd");
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"));
Paypal returns a timestamp of the following format:
yyyy-MM-ddTHH:mm:ssZ
And I don't quite know what to do with it...
How can I convert it to yyyy-MM-dd HH:mm:ss using my local timezone in php?
I'm tempted to preg_replace the mysterious letters, but something tells me there must a better way. There also appears to be 8 hours difference to my zone which I'm not sure how to substract.
Use DateTime class to do your magic.
$date = new DateTime('2012-09-09T21:24:34Z');
$date->format('Y-m-d'); # read format from date() function
You can use strtotime() to get a UNIX timestamp. From there you can do whatever you need: DateTime object, date(), etc.
Example with date():
echo date('r', strtotime('2012-09-10T10:00:00Z'));
strtotime() in PHP works great if you can provide it with a date format it understands and can convert, but for example you give it a UK date it fails to give the correct unix timestamp.
Is there any PHP function, official or unofficial, that can accept a format variable that tells the function in which format the date and time is being passed?
The closest I have come to doing this is a mixture of date_parse_from_format() and mktime()
// Example usage of the function I'm after
//Like the date() function but in reverse
$timestamp = strtotimeformat("03/05/2011 16:33:00", "d/m/Y H:i:s");
If you have PHP 5.3:
$date = DateTime::createFromFormat('d/m/Y H:i:s', '03/05/2011 16:33:00');
echo $date->getTimestamp();
You are looking for strptime, I think. you can use it to parse the date and then use mktime if you need a UNIX timestamp.
function strotimeformat($date, $format) {
$d = strptime($date, $format);
return mktime($d['tm_hour'], $d['tm_min'], $d['tm_sec'],
$d['tm_mon'], $d['tm_mday'], $d['tm_year']);
}
This will work with PHP 5.1 and onwards.
strtotime assumes it's a US date/time when using / as the separator. To get it to think it's a Euro date/time, use - or . as the date separator. You can change the /s to -s or .s with a simple str_replace()
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
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;
}
}