I am trying to extract the time from each field for now what i get is:
if i use the following code:
$row["real_stime"] = $row["start_time"];
$row["real_etime"] = $row["end_time"];
i get this output:
"real_stime":"2015-11-18 07:18:00","real_etime":"2015-11-18 20:18:00"
and if i use this code:
$row["formated_start_time"] = date("H:i",$row["start_time"]);
$row["formated_end_time"] = date("H:i",$row["end_time"]);
i get this output:
"formated_start_time":"00:33","formated_end_time":"00:33"
the output that i need is:
07:18,20:18
Use H for 24 hour format and h for 12 hour format
$row["formated_start_time"] = date("H:i",strtotime$row["start_time"]));
$row["formated_end_time"] = date("H:i",strtotime$row["end_time"]));
See this link for more information on formats
See Demo Here
use date('H:i:s')
this should be
date("G:i:s",$row["start_time"])
^
change to this
date("H:i:s",$row["start_time"]);
So final well-form code is
$row["formated_start_time"] = date("H:i:s",$row["start_time"]);
$row["formated_end_time"] = date("H:i:s",$row["end_time"]);
Related
Carbon::create($bulan)->lastOfMonth()->format('d')
I got this following code that search for the last day of the month with d format which return the string number "31", but i want to use the 31 as int how to convert it ?
EDIT
$tglA = Carbon::create($request->tglA)->startOfMonth()->format('Y-m-d');
$tglB = Carbon::create($request->tglB)->lastOfMonth()->format('Y-m-d');
$periodBulan = CarbonPeriod::create($tglA, '1 month', $tglB)->toArray();
$bulanInterval = [];
foreach($periodBulan as $bulan){
$temp = Carbon::create($bulan)->lastOfMonth()->format('d')->toDateTimeString();
array_push($bulanInterval, (int)$temp);
};
This is my code that i working on (already tried not using toDateTimeString still error could not be converted to int
If you get your value as a string, then you can cast it, like in this answer How do I convert a string to a number in PHP?
By simply using this code
$int = (int)$day;
in wich $day is your "31" as a string
Edit: Just checked that you already tried this solution
I have a really simple error and I don't get why stftime PHP function returns me an empty string while all seems right with the syntax and code.
I got the following code:
$next_high_tide_date = strtotime($next_high_tide_date);
// FROM '2021-05-22 00:26:00' TO (timestamp) 1621635960
$next_high_tide_date_day = strtoupper(strftime('%A %d %B', $next_high_tide_date));
// THIS RETURNS ME THE RIGHT DAY DATE
$next_high_tide_date_hour = strtoupper(strftime('%k H %M', $next_high_tide_date));
// THIS RETURNS ME AN EMPTY STRING
My $next_hight_tide_date value after the first declaration is equal to 1621643160 (timestamp).
I checked on a strftime checker online and my code is right, but in my php page, still returns me a empty string.
Anyone has an idea ?
Maybe you should use the date() function to format your timestamp.
$next_high_tide_date = "2021-05-22 00:26:00";
$next_high_tide_date = strtotime($next_high_tide_date);
$next_high_tide_date_day = strtoupper(date('l d F', $next_high_tide_date));
$next_high_tide_date_hour = strtoupper(date('H \H i', $next_high_tide_date));
try this
or use %H instead of %k
$next_high_tide_date = "2021-05-22 00:26:00";
$next_high_tide_date = strtotime($next_high_tide_date);
$next_high_tide_date_day = strtoupper(strftime('%A %d %B',$next_high_tide_date));
$next_high_tide_date_hour = strtoupper(strftime('%H H %M',$next_high_tide_date));
How can I force the date format to output:
12/12/2012, 1/10/2012, 1/5/2012
instead of
12/12/2012, 01/10/2012, 01/05/2012?
My code is the following:
$adatefrom = date_create($_POST['datefrom']);
$adateto = date_create($_POST['adateto']);
$adatefrom = date_format($adatefrom, 'd/m/Y');
$adateto = date_format($adateto, 'd/m/Y');
Please do note that I have to format the date AFTER posting it.
Have a look at the PHP built in date function here
You will find that your solution is as simple as this:
date('j/n/Y',strtotime($_POST['datefrom']));
The key things to note are the characters used in the first parameter.
j represents the day without leading zeros
n represents the month without leading zeros
There are many other options you have, just have a read through the documentation.
Please note that a simple search of 'PHP date' on Google would have found this solution for you
$adatefrom = date_create($_POST['datefrom']);
$adateto = date_create($_POST['adateto']);
$adatefrom = date_format($adatefrom, 'j/n/Y');
$adateto = date_format($adateto, 'j/n/Y');
you are welcome! ;)
Code:
$today_mem = date("d.m");
echo $today_mem; // -> 15.02
I need to transform dates like 15.02 into 15.2, also need to transform for ex. 07.02 into 7.2 and so on every day.
So the question is: how to delete firstdigit 0 from date and month.
Any short solutions?
You'll want to use:
$today_mem = date("j.n");
echo $today_mem; // -> 15.2
To remove the leading zeros. See more format modifiers at: php.net/date
Use j instead of d and n instead of m:
$today_mem = date("j.n");
Reference at the PHP doc site.
use
$today_mem = date("j")
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.