Currently I'm doing
$dayNames = array(
'Sun'=>'Sunday',
'Mon'=>'Monday',
'Tue'=>'Tuesday',
'Wed'=>'Wednesday',
'Thu'=>'Thursday',
'Fri'=>'Friday',
'Sat'=>'Saturday',
);
$weekday=$dayNames[$dayCode]; //where $dayCode is Sun, Mon...Sat
Is it the good way or there are any native PHP function for that ?
One possible approach:
$shortDay = 'Sun';
$fullDay = date('l', strtotime($shortDay)); // Sunday
Explanation: with strtotime, you essentially parse the given short dayname, constructing a new date based on it. With date('l') (it's lowercased L, not 1), you format it back into full textual representation of the day of the week.
Related
Assuming you have four numeric variables:
$Index=2; // 1-6, 2 is the second occurrence
$Dow=3; // 0-6, 3 is Wednesday
$Month=3; // 1-12, 3 is March
$Year=2016;
Is there a simple way to use those numbers to create or modify a DateTime object to represent that specific date? For example, with the variables above, it would be the specific date of the 2nd Wednesday in March of 2016.
I know that you can do this with strtotime or the DateTime constructor if you have it written out as a string already. There are other answers on here that deal with that.
But since we have numeric indexes, it seems stupid to create a function that converts numbers to a human readable string that then gets passed into a string parser to generate a date.
So the question is, is there a way to use strictly numeric variables with the DateTime object without having to first convert it to a human readable string to get a relative date like the 2nd(2) Wednesday(3) of March(3)?
I'm looking for something like:
$oDate = new DateTime();
$oDate->setFromWeekdayOfMonthIndex($Position,$DOWNum,$MonthNum,$Year);
UPDATE #1
Just to clarify the question a bit, this does what I want, but is a long/silly way of doing it:
$oMonthDate = new DateTime();
$oMonthDate->setDate($Year, $MonthNum, 1);
$oFormater = new NumberFormatter('en-US', NumberFormatter::SPELLOUT);
$oFormater->setTextAttribute(NumberFormatter::DEFAULT_RULESET,"%spellout-ordinal");
$String = $oFormater->format($Position).' ';
$String .= date('l', strtotime('Sunday +'.$DOW.' days')). ' of ';
$String .= $oMonthDate->format('F'). ' '.$Year;
$oFinalDate = new DateTime($String); // "second Tuesday of January 2016"
I have a PHP date in a database, for example 8th August 2011. I have this date in a strtotime() format so I can display it as I please.
I need to adjust this date to make it 8th August 2013 (current year). What is the best way of doing this? So far, I've been racking my brains but to no avail.
Some of the answers you have so far have missed the point that you want to update any given date to the current year and have concentrated on turning 2011 into 2013, excluding the accepted answer. However, I feel that examples using the DateTime classes are always of use in these cases.
The accepted answer will result in a Notice:-
Notice: A non well formed numeric value encountered......
if your supplied date is the 29th February on a Leapyear, although it should still give the correct result.
Here is a generic function that will take any valid date and return the same date in the current year:-
/**
* #param String $dateString
* #return DateTime
*/
function updateDate($dateString){
$suppliedDate = new \DateTime($dateString);
$currentYear = (int)(new \DateTime())->format('Y');
return (new \DateTime())->setDate($currentYear, (int)$suppliedDate->format('m'), (int)$suppliedDate->format('d'));
}
For example:-
var_dump(updateDate('8th August 2011'));
See it working here and see the PHP manual for more information on the DateTime classes.
You don't say how you want to use the updated date, but DateTime is flexible enough to allow you to do with it as you wish. I would draw your attention to the DateTime::format() method as being particularly useful.
strtotime( date( 'd M ', $originaleDate ) . date( 'Y' ) );
This takes the day and month of the original time, adds the current year, and converts it to the new date.
You can also add the amount of seconds you want to add to the original timestamp. For 2 years this would be 63 113 852 seconds.
You could retrieve the timestamp of the same date two years later with strtotime() first parameter and then convert it in the format you want to display.
<?php
$date = "11/08/2011";
$time = strtotime($date);
$time_future = strtotime("+2 years", $time);
$future = date("d/m/Y", $time_future);
echo "NEW DATE : " . $future;
?>
You can for instance output it like this:
date('2013-m-d', strtotime($myTime))
Just like that... or use
$year = date('Y');
$myMonthDay = date('m-d', strtotime($myTime));
echo $year . '-' . $myMonthDay;
Use the date modify function Like this
$date = new DateTime('2011-08-08');
$date->modify('+2 years');
echo $date->format('Y-m-d') . "\n";
//will give "2013-08-08"
I have data coming from the database in a 2 digit year format 13 I am looking to convert this to 2013 I tried the following code below...
$result = '13';
$year = date("Y", strtotime($result));
But it returned 1969
How can I fix this?
$dt = DateTime::createFromFormat('y', '13');
echo $dt->format('Y'); // output: 2013
69 will result in 2069. 70 will result in 1970. If you're ok with such a rule then leave as is, otherwise, prepend your own century data according to your own rule.
One important piece of information you haven't included is: how do you think a 2-digit year should be converted to a 4-digit year?
For example, I'm guessing you believe 01/01/13 is in 2013. What about 01/01/23? Is that 2023? Or 1923? Or even 1623?
Most implementations will choose a 100-year period and assume the 2-digits refer to a year within that period.
Simplest example: year is in range 2000-2099.
// $shortyear is guaranteed to be in range 00-99
$year = 2000 + $shortyear;
What if we want a different range?
$baseyear = 1963; // range is 1963-2062
// this is, of course, years of Doctor Who!
$shortyear = 81;
$year = 100 + $baseyear + ($shortyear - $baseyear) % 100;
Try it out. This uses the modulo function (the bit with %) to calculate the offset from your base year.
$result = '13';
$year = '20'.$result;
if($year > date('Y')) {
$year = $year - 100;
}
//80 will be changed to 1980
//12 -> 2012
Use the DateTime class, especially DateTime::createFromFormat(), for this:
$result = '13';
// parsing the year as year in YY format
$dt = DateTime::createFromFormat('y', $result);
// echo it in YYYY format
echo $dt->format('Y');
The issue is with strtotime. Try the same thing with strtotime("now").
Simply prepend (add to the front) the string "20" manually:
$result = '13';
$year = "20".$result;
echo $year; //returns 2013
This might be dumbest, but a quick fix would be:
$result = '13';
$result = '1/1/20' . $result;
$year = date("Y", strtotime($result)); // Returns 2013
Or you can use something like this:
date_create_from_format('y', $result);
You can create a date object given a format with date_create_from_format()
http://www.php.net/manual/en/datetime.createfromformat.php
$year = date_create_from_format('y', $result);
echo $year->format('Y')
I'm just a newbie hack and I know this code is quite long. I stumbled across your question when I was looking for a solution to my problem. I'm entering data into an HTML form (too lazy to type the 4 digit year) and then writing to a DB and I (for reasons I won't bore you with) want to store the date in a 4 digit year format. Just the reverse of your issue.
The form returns $date (I know I shouldn't use that word but I did) as 01/01/01. I determine the current year ($yn) and compare it. No matter what year entered is if the date is this century it will become 20XX. But if it's less than 100 (this century) like 89 it will come out 1989. And it will continue to work in the future as the year changes. Always good for 100 years. Hope this helps you.
// break $date into two strings
$datebegin = substr($date, 0,6);
$dateend = substr($date, 6,2);
// get last two digits of current year
$yn=date("y");
// determine century
if ($dateend > $yn && $dateend < 100)
{
$year2=19;
}
elseif ($dateend <= $yn)
{
$year2=20;
}
// bring both strings back into one
$date = $datebegin . $year2 . $dateend;
I had similar issues importing excel (CSV) DOB fields, with antiquated n.american style date format with 2 digit year. I needed to write proper yyyy-mm-dd to the db. while not perfect, this is what I did:
//$col contains the old date stamp with 2 digit year such as 2/10/66 or 5/18/00
$yr = \DateTime::createFromFormat('m/d/y', $col)->format('Y');
if ($yr > date('Y')) $yr = $yr - 100;
$md = \DateTime::createFromFormat('m/d/y', $col)->format('m-d');
$col = $yr . "-" . $md;
//$col now contains a new date stamp, 1966-2-10, or 2000-5-18 resp.
If you are certain the year is always 20 something then the first answer works, otherwise, there is really no way to do what is being asked period. You have no idea if the year is past, current or future century.
Granted, there is not enough information in the question to determine if these dates are always <= now, but even then, you would not know if 01 was 1901 or 2001. Its just not possible.
None of us will live past 2099, so you can effectively use this piece of code for 77 years.
This will print 19-10-2022 instead of 19-10-22.
$date1 = date('d-m-20y h:i:s');
Say it I get a date from MySQL table like 2012-03-31.
I pass this date to a Java application. So Java needs that date as 2012, 02, 31.
Firstly I explode date, get second element, subtract one from the month value. Then I implode three elements and create new date string.
public function convertToJavaDate($mysqlDate) {
$pieces = explode("-",$mysqlDate);
return $pieces[0].", ".($pieces[1]-1).", ".$pieces[2];
}
Is there a quicker or smarter way to do this ?
For java see http://docs.oracle.com/javase/1.4.2/docs/api/java/text/SimpleDateFormat.html#parse(java.lang.String, java.text.ParsePosition)
$date = strtotime('2012-03-31');
$javadata = date('Y, m, d', strtotime('-30 days', $date));
I have a problem where I need to handle dates where the month and day parts are optional. For example, the year will always be known but sometimes the day or month and day will be unknown.
In MySQL I can create a table with a date field and while I can't find any reference in the MySQL Manual it will accept the following as valid:
(YYYY-MM-DD format):
2011-02-10 // Current date
2011-02-00 // Day unknown so replaced with 00
2011-00-00 // Day and month unkown so replaced with 00-00
Test calculations from within the database work fine so I can still sort results easily. In the manual it says that month needs to be between 01 and 12, and day between 01 and 31 - but it does accept 00.
First question: Am I going to run into trouble using 00 in the month or day parts or is this perfectly acceptable?
Next question: Is there a PHP function (or MySQL format command) that will automatically format the following dates into the required format string?
2011 becomes 2011-00-00
2011-02 becomes 2011-02-00
Or do I need write a special function to handle this?
The following doesn't work:
<?php
$date = date_create_from_format('Y-m-d', '2011-00-00');
echo date_format($date, 'Y-m-d');
// Returns 2010-11-30
$date = date_create_from_format('Y-m-d', '2011-02-00');
echo date_format($date, 'Y-m-d');
// Returns 2011-01-31
?>
Third question: Is there a PHP function (or MySQL command) to format the dates for use in PHP?
Finally, is this the best approach? Or is there a 'best practise' method?
EDIT:
Here is what I'm currently doing:
A date field can accept a date in the format YYYY, YYYY-MM, or YYYY-MM-DD and before sending to the database it is processed in this function:
/**
* Takes a date string in the form:
* YYYY or
* YYYY-MM or
* YYYY-MM-DD
* and validates it
*
* Use date_format($date, $format); to reverse.
*
* #param string $phpDate Date format [YYYY | YYYY-MM | YYYY-MM-DD]
*
* #return array 'date' as YYYY-MM-DD, 'format' as ['Y' | 'Y-m' | 'Y-m-d'] or returns false if invalid
*/
function date_php2mysql($phpDate) {
$dateArr = false;
// Pattern match
if (preg_match('%^(?P<year>\d{4})[- _/.]?(?P<month>\d{0,2})[- _/.]?(?P<day>\d{0,2})%im', trim($phpDate), $parts)) {
if (empty($parts['month'])) {
// Only year valid
$date = $parts['year']."-01-01";
$format = "Y";
} elseif (empty($parts['day'])) {
// Year and month valid
$date = $parts['year']."-".$parts['month']."-01";
$format = "Y-m";
} else {
// Year month and day valid
$date = $parts['year']."-".$parts['month']."-".$parts['day'];
$format = "Y-m-d";
}
// Double check that it is a valid date
if (strtotime($date)) {
// Valid date and format
$dateArr = array('date' => $date, 'format' => $format);
}
} else {
// Didn't match
// Maybe it is still a valid date
if (($timestamp = strtotime($phpDate)) !== false) {
$dateArr = array('date' => date('Y-m-d', $timestamp), 'format' => "Y-m-d");
}
}
// Return result
return $dateArr;
}
So it pattern matches the input $phpDate where it must begin with 4 digits, then optionally pairs of digits for the month and the day. These are stored in an array called $parts.
It then checks if months or days exist, specifying the format string and creating the date.
Finally, if everything checks out, it returns a valid date as well as a format string. Otherwise it returns FALSE.
I end up with a valid date format for my database and I have a way of using it again when it comes back out.
Anyone think of a better way to do this?
I have a problem where I need to handle dates where the month and day parts are optional.
For example, the year will always be known but sometimes the day or month and day will be
unknown.
In many occasions, we do need such 'more or less precise' dates, and I use such dates as 2011-04-01 (precise), as well as 2011-04 (= April 2011) and 2011 (year-only date) in archives metadata. As you mention it, MySQL date field tolerates '2011-00-00' though no FAQs tell about it, and it's fine.
But then, I had to interface the MySQL database via ODBC and the date fields
are correctly translated, except the 'tolerated' dates (Ex: '2011-04-00' results empty in the resulting MySQL-ODBC-connected ACCESS database.
For that reason, I came to the conclusion that the MySQL date field could be converted in a plain VARCHAR(10) field : As long as we don't need specific MySQL date functions, it works fine, and of course, we can still use php date functions and your fine date_php2mysql() function.
I would say that the only case when a MySQL date field is needed
is when one needs complex SQL queries, using MySQL date functions in the query itself.
(But such queries would not work anymore on 'more or less precise' dates!...)
Conclusion : For 'more or less precise' dates,
I presently discard MySQL date field and use plain VARCHAR(10) field
with aaaa-mm-jj formated data. Simple is beautiful.
Since the data parts are all optional, would it be tedious to store the month, day, and year portions in separate integer fields? Or in a VARCHAR field? 2011-02-00 is not a valid date, and I wouldnt't think mysql or PHP would be excited about it. Test it out with str_to_time and see what kind of results you get, also, did you verify that the sorting worked right in MySQL? If the docs say that 1 through 31 is required, and it is taking 00, you might be relying on what is, in essence, a bug.
Since 2011-02-00 is not a valid date, none of PHP's formatting functions will give you this result. If it handled it at all, I wouldn't be surprised if you got 2001-01-31 if you tried. All the more reason to either store it as a string in the database, or put the month, day, and year in separate integer fields. If you went with the latter route, you could still do sorting on those columns.
I have also encountered this problem. I ended up using the PEAR Date package. Most date classes won't work with optional months or optional days, but the PEAR Date package does. This also means you don't need custom formatting functions and can use the fancy formatting methods provided by the Date package.
I have found this link in a textbook. This states that month and day values can be zero to allow for the possiblity of storing incomplete or unknown data
http://books.google.co.uk/books?id=s_87mv-Eo4AC&pg=PA145&lpg=PA145&dq=mysql+date+of+death+when+month+unknown&source=bl&ots=tcRGz3UDtg&sig=YkwpkAlDtBP1KKTDtqSyZCl63hs&hl=en&ei=Btf5TbL1NIexhAfkveyTAw&sa=X&oi=book_result&ct=result&resnum=8&ved=0CFMQ6AEwBw#v=onepage&q&f=false
If you pull your date in pieces from the database you can get it as if it's 3 fields.
YEAR(dateField) as Year, MONTH(dateField) as Month, DAY(dateField) as DAY
Then pushing those into the corresponding fields in the next bit of PHP will give you the result you're looking for.
$day = 0;
$month = 0;
$year = 2013;
echo $datestring;
$format = "Y";
if($month)
{
$format .= "-m";
if($day)
$format .="-d";
else
$day = 1;
}
else
{
$month = 1;
$day = 1;
}
$datestring = strval($year)."-".strval($month)."-".strval($day);
$date = date($format, strtotime($datestring));
echo $date; // "2013", if $month = 1, "2013-01", if $day and $month = 1, "2013-01-01"