Dates in PHP 4.4 - php

I am working for this association in my school and I am migrating web applications since, as of today, they are still running in PHP 4.4... But, I would like to quickly implement some fixes before changing to a newer PHP version.
One of these changes is related to dates comparison. The problem is I don't have strtotime and stuff like that, they all came in PHP 5.x.
I have to check if a timestamp is in one of two time intervals. In concrete:
I have a timestamp for when a person adhered to the association, school year goes from September 1st of year N to July 31st of year N+1, and I must check if the timestamp (and also the current time) lies
Between September 1st of year N and January 31st of year N+1, or
Between January 1st of year N+1 and July 31st of year N+1
So the question is what is the best way to do this? I came up with a solution: create my own timestamps and then compare. To create a timestamp for year N, I multiply the number of years since epoch time (N - 1970) by the number of seconds in a year ( 365.25 * 24 * 60 * 60 ). Then to that timestamp I add the number of seconds to September 1st, to January 1st, and so on. Finally I compare the whole. Isn't there a better way?

You are able to use PHP's strtotime function in PHP 4.4:
http://php.net/manual/en/function.strtotime.php
strtotime
(PHP 4, PHP 5, PHP 7)
strtotime — Parse about any English textual datetime description into a Unix timestamp
Here is an online example where you can run strtotime in 4.4.9:
http://sandbox.onlinephpfunctions.com/code/38559a58820d08105f746691bf338c565c1ed4e0

Related

Convert date format 1542575966120 18/11/2018 [duplicate]

This question already has answers here:
Convert one date format into another in PHP
(17 answers)
Closed 4 years ago.
I have a DB that shows when the user made the last login but then it shows 1542575966120. I wanted it to show so 18/11/2018 19:00
I tried using this in php
$intDate = "20". $ infologado ["lastlogin"];
$newDate = date ("d-m-Y", strtotime ($ intDate));
but I could not.
sorry for English
So as #Taha Paksu had mentioned, these numbers are a timestamp (seconds since 1 January 1970). Try this code:
$intDate = 1542575966120;
$newDate = date('d/m/Y H:i', $intDate/1000);
It is in miliseconds, date function accepts seconds, thus the division by 1000. Also no need to put it into strtotime, because this function is meant to convert string dates to... said numeric timestamps.
In your case, you can put $intDate = $infologado['lastlogin']; instead of first line to get the result dynamically from the database.
First of all, you need to learn what a timestamp is. The timestamp is a number which shows the seconds passed (or milliseconds, some include the milliseconds too) since epoch (01/01/1970). A general definition can be found here:
The Unix epoch (or Unix time or POSIX time or Unix timestamp) is the number of seconds that have elapsed since January 1, 1970 (midnight UTC/GMT), not counting leap seconds (in ISO 8601: 1970-01-01T00:00:00Z). Literally speaking the epoch is Unix time 0 (midnight 1/1/1970), but 'epoch' is often used as a synonym for 'Unix time'. Many Unix systems store epoch dates as a signed 32-bit integer, which might cause problems on January 19, 2038 (known as the Year 2038 problem or Y2038).
The converter on this page converts timestamps in seconds, milliseconds and microseconds to readable dates.
Taken from: https://www.epochconverter.com/ a tool which you can convert your dates to/from timestamps or vice versa.
Then to answer your question, the system saved the dates as a timestamp to the database to (probably) bypass the formatting errors on each different system that uses it.
Nevermind, TL;DR:
The number shows Sunday, 18 November 2018 21:19:26.120 when you give it to the timestamp converter I mentioned above. With PHP, you can use:
$unixTimestamp = 1542575966120;
$dt = DateTime::createFromFormat("U.u", $unixTimestamp / 1000);
var_dump($dt);
to convert to PHP DateTime class, then you can use it in your application.

strtotime skipping month when giving starting date

I'm trying to jump from month to month, starting from a specific timestamp, but when I get jump from August, September always gets skipped. Starting From August 31 (1346389200) and jumping 1 month:
strtotime('+1 month', 1346389200);
yields 1349067600 - which is October 1st.
I've read all about strtotime making mistakes if it doesn't have a starting date to calculate from, but what could the issue be with this?
Thanks
One month after 31 August is 31 September, but, because it not exists, php force the result to 1 Oktober.
So, you should force the current month on 1th day (of course if you want only year and month) :
strtotime('+1 month',strtotime(date("Y-m-1",1346389200)));
but if you use php >5.3 you can use more reliable DateTime class and methods.
You probably shouldn't use timestamps and strtotime for this comparison. You will introduce problems because of things like daylight savings, leap years, etc. Best to use DateTime and DateInterval classes/functions to do this in a more thorough manner.
http://php.net/manual/en/datetime.add.php

PHP DATE strtotime date "0001-01-01" [duplicate]

This question already has answers here:
Using strtotime for dates before 1970
(7 answers)
Closed 7 years ago.
Hello guys is there a way to convert the date 0001-01-1 to a time format? I am trying to to use the php function strtotime("0001-01-01"), but it returns false.
My goal is to count the days from the date: 0001-01-01 to: 2011-01-01.
Use DateTime class. strtotime returns a timestamp which in your case will be out of int bounds.
As Ignacio Vazquez-Abrams has commented, dates before the adoption of the Gregorian calendar are going to be problematic:
The Gregorian calendar was adopted at different times in different countries (anywhere from 1582 to 1929). According to Wikipedia, a few locales still use the Julian calendar.
Days needed to be "removed" to switch to the Gregorian calendar, and the exact "missing" dates differ between countries (e.g. the missing days in September 1752). Some countries tried to do a gradual change (by removing February 29 for a few years); Sweden switched back to the Julian calendar to "avoid confusion", resulting in a year with February 30.
Years were not always counted from January 1 (which has left its legacy in things like the UK tax year).
Michael Portwood's post The Amazing Disappearing Days gives a reasonable summary.
In short, you have to know precisely what you mean by "1 Jan 1 AD" for your question to make any sense. Astronomers use the Julian Day Number (not entirely related to the Julian calendar) to avoid this problem.
EDIT: You even have to be careful when things claim to use the "proleptic Gregorian calendar". This is supposed to be the Gregorian calendar extended backwards, but on Symbian, years before 1600 observe the old leap year rule (i.e. Symbian's "year 1200" is not a leap year, where in the proleptic Gregorian calendar it is).
I am afraid it wont work since strtotime changes string to timestamp, which have lowest value of 0, and its at 1970-01-01.. cant go lower than that..
date('Y-m-d', 0);
You cannot do that this way. Here is one option how to do this.
$date1= "2011-01-01";
$date2 = "2011-10-03";
//calculate days between dates
$time_difference = strtotime($date2) - strtotime($date1);
now you got time difference in seconds from wich you can get days, hours, minutes, seconds.
From the manual:
If the number of the year is specified in a two digit format, the values between 00-69 are mapped to 2000-2069 and 70-99 to 1970-1999. See the notes below for possible differences on 32bit systems (possible dates might end on 2038-01-19 03:14:07).
You may want to give 'DateTime::createFromFormat' a shot instead this.
Also note that you cannot go below 1901 on a 32-bit version of PHP.
You can't use the function strtotime because this function return a timestamp and the time stamp start from 1970 so i advise you to searsh for a different way

Convert PHP date to Excel?

Excel's date format is, I believe, the # of days since December 30, 1899. (Why? Because it's based on 1/1/1900 but they erroneously include a leap day in 1900, and it's one-based. So it's fine for the first couple of months in 1900 but then it goes wrong. 1/1/1901 in Excel is "367".)
So, how can I convert a date or timestamp from PHP (which is typically stored as a Unix timestamp, the # of seconds since 1/1/1970) to Excel?
In PHP 5.1.6?
Yeah, bet you didn't see that one coming. So I don't have any of the DateTime objects, date_create(), date_diff(), etc. functions available to me.
Is it even possible without simply recording the Excel day for 1/1/1970 and working from there? And is there a DST-proof version?
If I'm not mistaken, there are 86400 seconds in a day, and 25569 days between 30 Dec 1899 and 01 Jan 1970. So, to convert a Unix timestamp to an Excel date, the formula
ExcelDate = 25569 + UnixTS / 86400
ought to work. Of course, this formula is only correct for UTC. (Also, it ignores leap seconds.) For other timezones, if you know the offset (in seconds!) from UTC, you can just add it to the timestamp before using the formula above.

How to find if a date fits an interval in either PHP or MySQL?

Let's say I have a datetime, June 16 2011 at 7:00. I want to be able to check at, say, August 5 2011 at 7:00 and be able to tell that it is exactly a multiple of 1 day since the first date, whereas 7:01 would not count, since it is not an exact multiple.
Another test set: Let's say we have June 16 2011 at 7:00, and I want to check if a particular minute is within an interval of exactly 2 hours since then. So 9:00, 11:00, 13:00, etc. would count, but 9:30 and 10:00 would not. And this could continue for days and months - September 1 at 7:00 would still count as within every 2 hours. (And no, at the moment I don't know how I'm going to handle DST :D)
I thought about it for a moment and couldn't think of anything already existing in PHP or MySQL to do this easily but hell, it could, so I wanted to throw this up and ask before I start reinventing the wheel.
This is on PHP 5.1, sadly.
select *
from test
where datetimefield > '2011-06-16 07:00:00'
and
mod(timestampdiff(second,'2011-06-16 07:00:00',datetimefield),7200) = 0
This example will give you all the records greater than '2011-06-16 07:00:00' where the field is exactly a multiple of 2 hours.
Easiest would be to convert the date/time values into a unix timestamp and then simply do some subtraction/division:
2011-06-16 07:00:00 -> 1308229200
2011-08-05 07:00:00 -> 1312549200
2011-08-05 07:00:01 -> 1312549201
1312549200 - 1308229200 = 4320000 / 86400 = 50 (days)
1312549201 - 1308229200 = 4320001 / 86400 = 50.0000115...
So in other words:
if (($end_timestamp - $start_timestamp) % 864000)) == 0) {
... even multiple ...
}
Same would hold for the day/week comparisons. For months, this'll be out the window, since months aren't nice even figures to deal with.
MySQL Date functions:
http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html
You can use TIME() to get just the time part of a date. If the time parts are the same it is an exact multiple.
For the two hour thing, one way to do it would be to get the minute/seconds part of the date, make sure those are equal, then make sure that the hour parts of the dates are both even or both odd. For more complicated integer (e.g. 5) hour multiples, you can "fake" doing a mod by dividing the hour parts and checking if the result is an int.
You can compare two DateTime objects via diff() method. Result is a DateInterval object - you can check the exact number of days/hours/minutes between two dates.
It's useless to write your own algorithms if you can use built-in functionality.

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