I want to extract date without time from a single value from array result thaht I got.
I tried using array_slice but it didn't work.
My code..
$dateRows = $this->get('app')->getDates();
$dateRow = json_decode(json_encode($dateRows[0], true));
dump($dateRow[0]);die;
And I got result..
"2014-01-01 00:00:00"
And I want it to return just
"2014-01-01"
Very Simple, just use date_create() on your date & then format it using date_format() as follows -
$date = date_create("2014-01-01 00:00:00");
echo date_format($date,"Y-m-d");
So, in your case, it would be something like -
$date1 = date_create($dateRows[0]);
echo date_format($date1,"Y-m-d");
you can use strtotime function as well for getting formatted data
$a = "2014-01-01 00:00:00";
echo date('Y-m-d', strtotime($a));
Related
I have a variable is which the value coming is Date along with time in php. How do I convert it into a variable to get only the year? I do not need automatic updation but the format change is needed. Normal answers are giving it about date but my variable is containing time as well.
The format coming by now is 2017-12-11 4:06:37 and i need only 2017
Use like this:
<?php echo date('Y',strtotime('now'));?>
You can you simple DateTime function and date_formate() function for displaying separate year, month and date.
For that you have to first convert in Object of your current Date time string by using :
$date = new \DateTime('2017-12-11 4:06:37');
And then you can use date format function by using below code:
echo date_format($date, "Y"); //for Display Year
echo date_format($date, "m"); //for Display Month
echo date_format($date, "d"); //for Display Date
You can code like this (working perfectly):
$format = 'Y-m-d H:i:s';
$date = DateTime::createFromFormat($format, '2009-02-15 15:16:17');
echo "Format: $format; " . $date->format('Y') . "\n";
As mentioned by Himanshu Upadhyay, this is correct and the easiest way.
<?php
echo date('Y',strtotime('now'));
?>
But i would recommend you to read this here. You should really do actually!
By using DateTime class
$date = new \DateTime('2017-12-11 4:06:37');
echo $date->format('Y');
I have 2 variables stored into mysql:
campaign_date Format: d/m/Y
campaign_time Format: 24Hr
How could I concatenate them into one single variable like this:
2015-06-16T18:30
I tried with:
$new_datetime=$campaign_date.'T'.$campaign_time;
But it's not working
This should work for you:
(For the first date you have to change / to - so you can use date() and you can change the order of d/m/Y to Y-m-d, after that it's a simple concatenation with the time at the end)
<?php
$campaign_date = "16/12/2014"; //Data from DB
$campaign_time = "18:00"; //Data from DB
echo $new_datetime = date("Y-m-d", strtotime(str_replace("/", "-", $campaign_date))) . "T" . date("H:i", strtotime($campaign_time));
?>
Output:
2014-12-16T18:00
Try it,i tested it myself.
$db_date = date("Y-m-d",strtotime($db_date));
$db_time = date("h:i:s",strtotime($db_time));
echo $db_date.'T'.$db_time;
I need to get only date from datetime variable. I was using this code. It was working fine but now I am facing problem when I'm trying to store only date in mysql please. Take a look at the code below and please tell me what I'm missing,
$date_time = "11-Dec-13 8:05:44 AM"
From the date I got from user input, I need to save only date in att_date variable.
$arr = explode("/", $date_time);
$arr2 = explode(" ", $arr[2]);
$att_date = $arr2[0].'-'.$arr[0].'-'.$arr[1];
Here is the solution:
$date_time = "11-Dec-13 8:05:44 AM";
$new_date = date("Y-m-d H:i:s",strtotime($date_time));
Even if you just use "Y-m-d" it will work instead of using "Y-m-d H:i:s",please practice below example:
$datetime = '2022-10-11 06:38:02';
$getOnlyDate = date('Y-m-d',strtotime($datetime));
echo $getOnlyDate; // output: 2022-10-11
You can get the date from the datetime variable using the
Date();
function from mysql using php query, it automatically extracts the date from it. You can learn it here.
I tried the following code to change the dateformat from dmy to ymd, but when using i got wrong dates.
My code
$sdate11=date("Y-m-d", strtotime($_POST["txtstartdates"]) );
$sdate111=date("Y-m-d", strtotime($_POST["txtenddates"]) );
dates inserted were
30-05-2013 and 31-05-2013
the date it returned was
2035-11-03 and 2036-11-02
could you please help me to find what was the problem here and solve it
Thank you.
Try with split like
$a = split('-',$_POST["txtstartdates"]);
or you can use explode even like
$a = explode('-',$_POST["txtstartdates"]);
$my_new_date = $a[2].'-'.$a[1].'-'.$a[0];
Here strtotime will not work for the format dd-mm-yyyy
You might use DateTime for that:
$date = DateTime::createFromFormat('d-m-Y', $_POST['txtstartdates']);
echo $date->format('Y-m-d');
$date = DateTime::createFromFormat('d-m-Y', $_POST['txtenddates']);
echo $date->format('Y-m-d');
Can't you use the DateTime object to convert the date into the format you want?
$DateTime = new DateTime($_POST['FIELD']);
echo $DateTime->format('Y-m-d');
Your code seems correct , I dont know why its not working for you Can you try below
I have made the '' from "" only
$sdate11=date('Y-m-d', strtotime($_POST['txtstartdates']));
$sdate111=date('Y-m-d', strtotime($_POST['txtenddates']));
You can do it using single line as:
$show_date = DateTime::createFromFormat('d-m-Y', $dateInput)->format('Y-m-d');
I'm having date 20/12/2001 in this formate . i need to convert in following format 2001/12/20 using php .
$var = explode('/',$date);
$var = array_reverse($var);
$final = implode('/',$var);
Your safest bet
<?php
$input = '20/12/2001';
list($day, $month, $year) = explode('/',$input);
$output= "$year/$month/$day";
echo $output."\n";
Add validation as needed/desired. You input date isn't a known valid date format, so strToTime won't work.
Alternately, you could use mktime to create a date once you had the day, month, and year, and then use date to format it.
If you're getting the date string from somewhere else (as opposed to generating it yourself) and need to reformat it:
$date = '20/12/2001';
preg_replace('!(\d+)/(\d+)/(\d+)!', '$3/$2/$1', $date);
If you need the date for other purposes and are running PHP >= 5.3.0:
$when = DateTime::createFromFormat('d/m/Y', $date);
$when->format('Y/m/d');
// $when can be used for all sorts of things
You will need to manually parse it.
Split/explode text on "/".
Check you have three elements.
Do other basic checks that you have day in [0], month in [1] and year in [2] (that mostly means checking they're numbers and int he correct range)
Put them together again.
$today = date("Y/m/d");
I believe that should work... Someone correct me if I am wrong.
You can use sscanf in order to parse and reorder the parts of the date:
$theDate = '20/12/2001';
$newDate = join(sscanf($theDate, '%3$2s/%2$2s/%1$4s'), '/');
assert($newDate == '2001/12/20');
Or, if you are using PHP 5.3, you can use the DateTime object to do the converting:
$theDate = '20/12/2001';
$date = DateTime::createFromFormat('d/m/Y', $theDate);
$newDate = $date->format('Y/m/d');
assert($newDate == '2001/12/20');
$date = Date::CreateFromFormat('20/12/2001', 'd/m/Y');
$newdate = $date->format('Y/m/d');