PHP 6 digit date code display in human readable format - php

Question

I have a set of dates that are formatted like this...
197402
192201
184707
The first four digits represents the year and the remaining two the month. I am trying to output these in this format
February 1974
January 1922
July 1847
I have tried passing it to the date function like this...
echo date ('F Y', 197402)
But this is giving me January 1970 everytime so I assume I have misunderstood how the date function works, can anyone help?

Answer 1

You're getting "January 1970" as an output, because you tried to create a date from the timestamp 197402, which is seconds from January 1st, 1970. If you output the full string from that (with seconds and whatnot), you'll see it's a valid timestamp, producing an actual date, but they all end up in the start of January 1970, see this online demo.
That format, YYYYMM, isn't a recognizable format for most functions. You need to split it up, if you know the format will be in that way - and use that data instead. You can use substr() to split the string, and then convert the numerical month to the string associated with that month, with the help of date() and mktime() (since you just specify the year and month).
The following snippet
$arr = [197402, 192201, 184707];
foreach ($arr as $v) {
$year = substr($v, 0, 4);
$month = substr($v, 4, 2);
echo date("F Y", mktime(0, 0, 0, $month, 0, $year))."<br />"; // mktime() produces a valid timestamp based on just month and year
// Alternatively, drop mktime() and use strtotime() and create from a standard format,
// while specifying a date in the month (which won't matter to the output)
// echo date("F Y", strtotime("$month/01/$year"))."<br />";
}
will output
February 1974
January 1922
July 1847
Alternatively, you can use the DateTime class (which is a lot simpler to work with), and create from a given format with date_create_from_format()
foreach ($arr as $v) {
echo date_create_from_format('Yh', $v)->format('F Y')."<br />";
}
This will generate the same output as above.
References
http://php.net/substr
http://php.net/mktime
http://php.net/date
http://php.net/datetime.createfromformat

Answer 2

I'd use the DateTime class, you can create from a specific format, and then output to another.
As pointed out in the comments below, you also need to set the day to the first of the month, otherwise you'll get undesired results if the current day is greater than the number of days in the given month.
echo DateTime::createFromFormat('Ymd', 19470201)->format('F Y');

Answer 3

As you are passing an int to date, it is considering it as a Unix Timestamp.
To create a date object from a predefined format, use DateTime::createFromFormat.
echo DateTime::createFromFormat('Ym',198403)->format('F Y');
results in
March 1984

Answer 4

You'd need to add a day to it, e.g. just add "01" and then use strtotime to convert that into a unix timestamp, as the date() function expects a timestamp as the parameter.
e.g.
echo date('F Y', strtotime("19220101"));

Answer 5

You have text representation of dates in a non-standard format. First you have to parse them and convert them to timestamps (the number of seconds since Jan 1, 1970 00:00:00 UTC). The PHP function date() can work only with timestamps.
The best approach (as of 2017) is to use the DateTime PHP class for date & time processing:
foreach (array('197402', '192201', '184707') as $text) {
$date = DateTime::createFromFormat('Ym', $text);
echo($date->format('F Y')."\n");
}
The method DateTime::createFromFormat() parses a string using the given format and creates a new DateTime object if the parsing succeeds. It is the OOP equivalent of strtotime() but smarter (because it can get hints about what date components to search in the input string.)
The method DateTime::format() produces the text representation of a date using the provided format. It is the OOP equivalent of date().
The OOP approach (the DateTime* classes) is recommended (and better than the procedural approach) because has built-in support for timezones (the procedural date-time functions lack it.)