Hi I am struggling with converting a string:
$file = '150220';
into a mysql date format.
I have tried this
$newDate = date("Y-m-d", strtotime($file));
But it returns the following 2015-02-24 (24 being today's date)
See: http://php.net/manual/en/datetime.createfromformat.php
So you can use this:
$dateTime = DateTime::createFromFormat('ymd', $file);
$newDate = $dateTime->format('Y-m-d');
strtotime is a wonderful function. It works when you provide YYYY-MM-DD format but not with YY-MM-DD.
EDIT: As said in comments, you have to "turn the string into a valid ISO8601 Notations format by prepending '20' - this is understood by strtotime as specified in the docs "
This will work for the next 85 years with 64 bits systems (Until 2038 on 32 bit systems)
$file = '150220';
$newDate = date("Y-m-d", strtotime("20$file"));
You have to be careful with strtotime because it returns false on error which will make date used today's date.
EDIT: if you want to -1 : first, this code works and it explains why op's code does not work, then consider using the "Add comment" button. Thanks.
Related
How to convert this (in ISO8601 format): 2014-03-13T09:05:50.240Z
To this (in MySQL DATE format): 2014-03-13
in php?
try this
$date = '2014-03-13T09:05:50.240Z';
$fixed = date('Y-m-d', strtotime($date));
The complete date function documentation can be found here: http://php.net/manual/en/function.date.php
The PHP function "strtotime" does nothing else then converting your timestring into an unix timestamp.
Hope I could help :)
P.s.:
Just in case strtotime will return 0 try using this:
$date = '2014-03-13T09:05:50.240Z';
$fixed = date('Y-m-d', strtotime(substr($date,0,10)));
Since PHP 5.2.0 you can do it using OOP and DateTime() as well (of course if you prefer OOP):
$now = new DateTime("2014-03-13T09:05:50.240Z");
echo $now->format('Y-m-d'); // MySQL datetime format
There is no reason to use the inefficient time functions. The most efficient way is to simply extract the first 10 characters:
substr($date,0,10)
People, that are really coding for year ≥10000, can use:
substr($date,0,strpos($date,"T"))
Simply convert datetime description into a Unix timestamp using with strtotime and then five format using Date Formats
Try it will surely work for you.
$date = '2014-03-13T09:05:50.240Z';
$fixed = date('Y-m-d', strtotime($date));
For those using Carbon (php library), the parse() works quite well:
Carbon::parse($date)
https://carbon.nesbot.com/docs/
Today I have published an interitty/utils package that deals with, among other things, the ISO-8601 format and perhaps all permutations of this standard.
I hope it will help you too.
$dateTimeFactory = new Interitty\Utils\DateTimeFactory();
$dateTime = $dateTimeFactory->createFromIso8601('1989-12-17T12:00:00Z');
I have a string that is a 10 digit date/time representation in the format of yymmddhhmm, e.g:
1304282240
is the 28th of April, 2013 22:40 (28/04/13 in the DD/MM/YY format)
I need to convert this into a valid PHP timestamp but I'm getting stuck using the Date and strtotime functions, and not sure if that is the best approach in any case.
Update: I've added these lines to my code:
$date = DateTime::createFromFormat('ymdHi', '1304282240');
echo $date->format('m/d/Y h:i:s A');
appears to work
I believe you want something like this:
$date = DateTime::createFromFormat('ymdHi', '1304282240');
I have been looking online for this answer and have come up empty...I am extremely tired so I thought I would give this a go....
I have a variable that has a date from a textbox
$effectiveDate=$_REQUEST['effectiveDate'];
What I am trying to do is take this date and add the current time
date('Y-m-d H:i:s', strtotime($effectiveDate))
When I echo this out I get 1969-12-31 19:00:00
Is this possible? Can someone point me in the right direction?
I found a solution to my problem....
$currentDate = date("Y-m-d");
$currentTime = date("H:i:s");
$currentDate = date("Y-m-d H:i:s", strtotime($currentDate . $currentTime));
echo $currentDate;
This takes a date from variable in one format and takes the date from another variable in another format and puts them together :)
Thanks everyone for their time.....
DateTime::createFromFormat
would also work but only if you have PHP 5.3 or higher...(I think)
The effectiveDate string is not in a format that strtotime recognizes, so strtotime returns false which is interpreted as 0 which causes the date to be displayed as January 1, 1970 at 00:00:00, minus your time zone offset.
The result you see is caused by the entered date not being in a format recognised by strtotime. The most likely case I can think of without knowing the format you used is that you used the US order of putting the month and day the wrong way around - this confuses strtotime, because if it accepts both then it can't distinguish February 3rd and March 2nd, so it has to reject US-formatted dates.
The most reliable format for strtotime is YYYY-MM-DD HH:ii:ss, as it is unambigous.
The date is just a timestamp, it is not object-oriented and i don't like it.
You can use the DateTime object.
The object-oriented best way is:
$effectiveDate=$_REQUEST['effectiveDate'];
// here you must pass the original format to pass your original string to a DateTimeObject
$dateTimeObject = DateTime::createFromFormat('Y-m-d H:i:s', $effectiveDate);
// here you must pass the desired format
echo $dateTimeObject->format('Y-m-d H:i:s');
$data['user']['time'] = '2011-03-07 00:33:45';
how can we add 1 year to this date ?
something like $newdata = $data['user']['time'] + 1 year ?
or
$newdata = 2012-03-07 00:33:45
Thanks
Adam Ramadhan
strtotime() is the function you're looking for:
$data['user']['seal_data'] = date('Y-m-d H:i:s', strtotime('+1 year', strtotime($data['user']['time'])));
First, you have to convert the MySQL datetime to something that PHP can understand. There are two ways of doing this...
Use UNIX_TIMESTAMP() in your query to tell MySQL to return a UNIX timestamp of the datetime column.
SELECT whatever, UNIX_TIMESTAMP(myTime) AS 'myUnixTime' FROM myTable;
Use DateTime::createFromFormat to convert your string time to something PHP can understand.
$date = DateTime::createFromFormat('Y-m-d H:i:s', $data['user']['time']);
Once that is done, you can work with the time... Depending on the method you used above, you can use one of the following.
If you have a unix timestamp, you can use the following to add a year:
$inAYear = strtotime('+1 year', $data['user']['unixTime']);
If you have a DateTime object, you can use the following:
$inAYear = $date->add(new DateInterval('P1Y'));
Now, to display your date in a format that is respectable, you must tell PHP to return a string in the proper format.
If you have a unix timestamp, you can use the following:
$strTime = date('Y-m-d H:i:s', $inAYear);
If you have a DateTime object, you can use the following:
$strTime = $inAYear->format('Y-m-d H:i:s');
Alternatively, if you don't want to deal with all of that, you can simply add one year when you query.
SELECT whatever, DATE_ADD(myTime, INTERVAL 1 YEAR) AS 'inAYear' FROM myTable;
Current (2017) Practice is to use DateTime
This question is top on a google search for "php datetime add one year", but severely outdated. While most of the previous answers will work fine for most cases, the established standard is to use DateTime objects for this instead, primarily due strtotime requiring careful manipulation of timezones and DST.
TL;DR
Convert to DateTime: $date = new DateTime('2011-03-07 00:33:45', [user TZ]);
Use DateTime::modify: $date->modify('+1 year');
Format to needs.
Change the timezone with DateTime::setTimezone from the list of supported timezones: $date->setTimezone(new DateTimeZone('Pacific/Chatham'));
Convert to string with DateTime::format: echo $date->format('Y-m-d H:i:s');
Following this pattern for manipulating dates and times will handle the worst oddities of timezone/DST/leap-time for you.
Just remember two final notes:
Life is easier with your system timezone set at UTC.
NEVER modify the system timezone outside of configuration files.
I've seen too much code that relies on date_default_timezone_set. If you're doing this, stop. Save the timezone in a variable, and pass it around your application instead, please.
More Reading
How to calculate the difference between two dates using PHP?
Convert date format yyyy-mm-dd => dd-mm-yyyy
PHP - strtotime, specify timezone
I think you could use strtotime() to do this pretty easily. Something like:
$newdata = date('c', strtotime($data['user']['time'] . ' +1 year'));
Though the 'c' format string isn't the same as your input format. You could consult date()'s docs for how to construct the correct one.
'Y-m-d H:i:s' — as Tim Cooper suggests — looks correct.
This should do the trick (not tested).
$data = "2011-03-07 00:33:45";
echo 'Original date +1 year: ' . date('Y-m-d H:i:s', strtotime(date("Y-m-d H:i:s", strtotime($data)) . " +1 year"));
First-of-all if your date format is separated by a slash (/), like '2019/12/31' then you should convert it in dash (-) format, like '2019-12-31', to do so use str_replace() function.
$string = str_replace('/', '-', '2019/12/31'); //output: 2019-12-31
To add time/day/month/year do not use strtotime() function, because it can't add a time which is beyond year 2038.
So here I would prefer to use DateTime() function.
$string = '2000-01-01';
$date = new DateTime($string);
$date->add(new DateInterval('P60Y5M2DT6H3M25S')); //60 Years 5 Months 2 Days 6 Hours 3 Minutes 25 Seconds
echo $date->format('Y-m-d H:i:s'); //output: 2060-06-03 06:03:25
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;
}
}