I need to check if entered date is between 2019-09-01 - 2019-12-31
I can do this as follows: $koodi is user input
$pattern1 = "/^2019-(09|11)-([0-2][0-9]|30)$/";
$pattern2 = "/^2019-(10|12)-([0-2][0-9]|3[0-1])$/";
if(preg_match($pattern1, $koodi) || preg_match($pattern2, $koodi)) {
echo "<code>$koodi</code> ok!<br>\n";
}
else {
echo ("<code>$koodi</code> NOT ok!<br>\n");
}
I was trying to make those two conditions into single regex statement, is that possible and if so how?
I tried:
$pattern = "/^2019-(09|11)-([0-2][0-9]|30)$ | ^2019-(10|12)-([0-2][0-9]|3[0-1])$/";
Did not work, neither the following where i tried to put parentheses around conditions:
$pattern = "/(^2019-(09|11)-([0-2][0-9]|30)$) | (^2019-(10|12)-([0-2][0-9]|3[0-1])$)/";
Please don't use a regex to do that, what if the dates change or what if the next developer has to come and work on this and figure out what your doing?
According to this article you can check if a date is between 2 dates by doing something like this.
<?php
$currentDate = date('Y-m-d');
$currentDate = date('Y-m-d', strtotime($currentDate));
$startDate = date('Y-m-d', strtotime("01/09/2019"));
$endDate = date('Y-m-d', strtotime("01/10/2019"));
if (($currentDate >= $startDate) && ($currentDate <= $endDate)){
echo "Current date is between two dates";
}else{
echo "Current date is not between two dates";
}
as for why your patterns didn't work its because you have a space around the pipe in the middle and you may possibly need to wrap the whole thing in brackets. You also have the $ half way through the regex which is matching the whole string, I would usually only have it at the end, like this: -
^(regex1|regex2)$
I haven't written the correct version in case your tempted to use it, (please use the date objects method)
Related
I want to display a text if the date in the database matches today's date.
$userdate = date("m/d/Y", strtotime($rr['last_login']));
$today = date("m/d/Y");
if ($userdate==$today){
echo "<test>";
}
Even if it is today's date, the records never echo out the string.
Interestingly, if I change it to
if ($userdate!=$today){
it also does not display the <test>.
My bad! I was not trying to echo out any html thing. The < brackets were just random characters. I changed it to "test" and now it works. Sorry for the < > characters causing confusion!
Since this is PHP, one would assume you're looking at the result in a browser, which will interpret the "<test>" as an HTML tag, and thus it will ignore it. Try changing the "echo" to simply echo "test<br>";
Also, if you are always dealing with a database, you could also use this:
if (substr($rr['last_login'], 0, 10) == date('Y-m-d'))
echo "test<br />\n";
i try to find out if a Timeperiod is inside a timeperiod. I have my reference time period and my comparative time period.
Let me make an example:
Time period A (reference) goes from 1.1.2014 to 1.2.2014 (tt.mm.yyyy).
Time period B (comparative) goes from 1.4.2014 to 1.5.2014.
=> This would be totaly ok.
Time period C (reference) goes from 1.1.2014 to 1.3.2014
Time period D (comparative) goes from 1.2.2014 to 1.5.2014.
=> Not ok because D is in C.
I hope you get what i want. I tried to make serval < = > if actions but this starts to get to huge and slow. Maybe there is a faster ways to do so.
Also, is MySQL able to do such things?
you can try this with php timestamp
$reference_start_date = "1.1.2014";
$reference_end_date = "1.2.2014";
$comparative_start_date = "1.4.2014";
$comparative_end_date = "1.5.2014 ";
$reference_start_time = strtotime(str_replace(".", "-", $reference_start_date);
$reference_end_time = strtotime(str_replace(".", "-", $reference_end_date);
$comparative_start_time = strtotime(str_replace(".", "-", $comparative_start_date);
$comparative_end_time = strtotime(str_replace(".", "-", $comparative_end_date);
if($comparative_start_time>$reference_start_time && $comparative_start_time<$reference_end_time)
{
echo "Not OK";
}
else if($comparative_end_time>$reference_start_time && $comparative_end_time<$reference_end_time)
{
echo "Not OK";
}
else if($comparative_start_time<$reference_start_time && $comparative_end_time>$reference_end_time)
{
echo "Not OK";
}
else
{
echo "OK";
}
you can do like below:
Check Reference_start >= comparative_start && Reference_end < comparative_end, If this condition become true than your time will be overlapped.
If you have a reference period (having startDate and endDate) and you have a comparative period, then you can have this where clause in MySQL:
where ((reference.startDate > comparative.endDate) or reference.endDate < comparative.startDate)
which would be true if the two periods have no intersection.
Assuming you have your dates give in UTC it is really simple to compare two date ranges. There are 5 specific cases that could happen:
11111......
......22222
..11111.....
.....22222..
...11111....
...22222....
.....11111..
..22222.....
......11111
22222......
Only the first and the last one are the ones you are looking for. It's easy to construct an if query of it and negate it:
if (!($dateRange1End <= $dateRangeStart2 && $dateRange2End <= $dateRange1Start))
// NOT OKAY
else
// OKAY
Test case scenario - User clicks on one of two links: 2012/10, or 2012/10/15.
I need to know whether the DAY is specified within the link. I am already stripping the rest of the link (except above) out of my URL, am I am passing the value to an AJAX request to change days on an archive page.
I can do this in either JS or PHP - is checking against the regex /\d{4}\/\d{2}\/\d{2}/ the only approach to seeing if the day was specified or not?
You can also do this if you always get this format: 2012/10 or 2012/10/15
if( str.split("/").length == 3 ) { }
But than there is no guaranty it will be numbers. If you want to be sure they are numbers you do need that kind of regex to match the String.
You could explode the date by the "/" delimiter, then count the items:
$str = "2012/10";
$str2 = "2012/10/5";
echo count(explode("/", $str)); // 2
echo count(explode("/", $str2)); // 3
Or, turn it into a function:
<?php
function getDateParts($date) {
$date = explode("/", $date);
$y = !empty($date[0]) ? $date[0] : date("Y");
$m = !empty($date[1]) ? $date[1] : date("m");
$d = !empty($date[2]) ? $date[2] : date("d");
return array($y, $m, $d);
}
?>
I would personally use a regex, it is a great way of testing this sort of thing. Alternatively, you can split/implode the string on /, you will have an array of 3 strings (hopefully) which you can then test. I'd probably use that technique if I was going to do work with it later.
The easiest and fastest way is to check the length of the string!
In fact, you need to distinguish between: yyyy/mm/dd (which is 10 characters long) and yyyy/mm (which is 7 characters).
if(strlen($str) > 7) {
// Contains day
}
else {
// Does not contain day
}
This will work EVEN if you do not use leading zeros!
In fact:
2013/7/6 -> 8 characters (> 7 -> success)
2013/7 -> 6 characters (< 7 -> success)
This is certainly the fastest code too, as it does not require PHP to iterate over the whole string (as using explode() does).
I've been googling a bit, but I can't figure out what keywords to use.
I'm saving the users date of birth, and I want to make sure the format is YYYY-MM-DD.
I'm thinking something like:
if(!ctype_digit(str_replace("-", "", $dob)) || strlen(str_replace("-", "", $dob)) != 8)
{
echo "Incorrect format: date of birth";
}
For this, I need to use strlen() to only replace three - chars. If it's more or less than 3, then echo incorrect format. How do I achieve this? Or is there a better way?
How about using regex?
if ( !preg_match( "/^[0-9]{4}-[0-9]{2}-[0-9]{2}$/", $dob) )
{
echo "Invalid date.";
}
To find many examples, search for "validating dates" or, in your case, to find php examples, try "php validate date format" -- you'll see most examples are solved with regular expressions, which is your ticket this this sort of thing. Reg Expressions can help you validate dates, email address formats, inputs for passwords that meet minimal requirements (e.g. at least 8 characters, at least one numeric and one capital letter, etc.)
Here's one such example:
if(preg_match('/\d{4}-\d{2}-\d{2}/',$date)){
// good date
}else{
// bad date
}
You should also check that the date is an actual date; not just that it looks like an actual date.
if (!preg_match("/^[0-9]{4}-[0-9]{2}-[0-9]{2}$/D", $dob))
{
echo 'Invalid date';
}
else
{
list($year, $month, $day) = explode('-', $dob);
if (!checkdate((int) $month, (int) $day, (int) $year))
{
echo 'Invalid date';
}
}
You don't want to accept 1988-06-31 after all.
im trying to validate a date to see if it matchs the mysql format
this is the code
$match = "/^\d{4}-\d{2}-\d{2} [0-2][0-3]:[0-5][0-9]:[0-5][0-9]$/";
$s = $this->input->post("report_start"). " " . $this->input->post("report_start_time").":00";
$e = $this->input->post("report_end"). " " . $this->input->post("report_end_time").":59";
if($this->input->post("action") != "")
{
echo trim($s). " => " . preg_match($match, trim($s));
echo "<br>";
echo trim($e). " => " . preg_match($match, trim($e));
}
the date format goes into $s and $e are
$s = 2011-03-01 00:00:00
$e = 2011-03-01 23:59:59
and they both return false (0).
i tested the pattern on http://www.spaweditor.com/scripts/regex/index.php and it returns true (1)
http://pastebin.com/pFZSKYpj
however if i manual inter the date strings into preg_match like
preg_match($match, "2011-03-01 00:00:00")
it works.
i have no idea what im doing wrong
======================
now that i think about it, i only need to validate the houre:min part of the datetime string.
im manually adding the seconds and the date is forced by a datepicker and users cant edit it
You're making your work harder that it needs to be. In php there are many date handling functions that mean you don't have to treat dates like strings. So, rather than test that your input dates are in the correct format, just insist on the correct format:
$adate= date_create('January 6, 1983 1:30pm'); //date format that you don't want
$mysqldate= $adate->format("Y-m-d h:i:s");//date format that you do want
There are also functions to check that a date is a real date, like checkdate.
ok heres wat i did.
since im forcing the date format and the ending seconds of the time part
i just validated the hour:mini part using "/^2[0-3]|[01][0-9]:[0-5][0-9]$";
and if that returns true i put everything together end reconstructed the final datetime string
$match = "/^2[0-3]|[01][0-9]:[0-5][0-9]$/";
$s_d = $this->input->post("report_start");
$s_t = $this->input->post("report_start_time");
$e_d = $this->input->post("report_end");
$e_t = $this->input->post("report_end_time");
if($this->input->post("action") != "")
{
if(
( preg_match($match , trim($s_d." ".$s_t.":00")) )
&& ( preg_match($match , trim($e_d." ".$e_t.":59")) )
)
{
$r = $this->model_report->client_hours_logged(array($s,$e));
$data['report'] = $r;
var_dump($r);
//$this->load->view("report/client_hours_per_client",$data);
}
}
Watch out:
[0-2][0-3] is not a good regex for hour values - it will match 01, 12, 23 and others, but it will fail 04 through 09 and 14 through 19.
Better use (2[0-3]|[01][0-9]) instead.
I use this to validate a 'Y-m-d H:i:s' format date string:
match = '/^[12][0-9]{3}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[01]) ([01][0-9]|2[0-3]):[0-5][0-9]:[0-5][0-9]$/';
You could use strtotime and date to parse and format the date properly.
Why not just simply force the date into the format you want:
$e = '2011-03-01 00:00:00';
$mysqlFormat = date('Y-m-d H:i:s', strtotime($e));
Also, there is a bit of an error in your regex [0-2][0-3]:[0-5][0-9]:[0-5][0-9] will only match the hours of 00,01,02,03,10,11,12,13,20,21,22,23 so it will never match 4am, or 3pm among others. That aside I looked over your RegEx and I don't see any problems with it matching the test cases you've offered. I would check to make sure there is not extra whitespace on either side of date string with trim().
I concur with Tim : MySQL behaves in quirks mode and always tries to go easy on DATE and DATE_TIME column types. You can omit certain parts of your input and it still will try to compensate and achieve that goal successfully to some degree... That's why, most numbers your Reg-ex considers as invalid, MySQL will accept as valid.