How to set a value for the input type 'datetime-local'?

Question

I tried this:

\" class=\"date\" name=\"start\" REQUIRED>

H

Answer 1

I don't know exacly what is in $row['Time'] but it should be as follows:

Definition

A valid date-time as defined in RFC 3339 with these additional qualifications:

• the literal letters T and Z in the date/time syntax must always be uppercase
• the date-fullyear production is instead defined as four or more digits representing a number greater than 0

Examples

• 1990-12-31T23:59:60Z
• 1996-12-19T16:39:57-08:00

---

Solution

To create RFC 3339 format in PHP you can use:

echo date('Y-m-d\TH:i:sP', $row['Time']);

or in another way:

echo date("c", strtotime($row['Time']));  

or if you prefer objective style:

echo (new DateTime($row['Time']))->format('c');

---

In your code

So in your code it would look as follows:

<input type="datetime-local"  value="<?php echo date('Y-m-d\TH:i:sP', $row['Time']); ?>" class="date" name="start" REQUIRED>

or

<input type="datetime-local"  value="<?php echo date("c", strtotime($row['Time'])); ?>" class="date" name="start" REQUIRED>

---

Manual

• More informations can be found here

• PHP date Manual

• PHP DateTime Manual

Answer 2

The answer of Karol Gasienica is a great explanation but somehow did not work for me even in their replies

date('Y-m-d\TH:i:s', $row['Time']); //Gives me 1970-01-01 00:00
date('Y-m-d\TH:i:sP', $row['Time']); //Gives me no display
date("c", strtotime($row['Time'])); //No display too

What worked for me is this

$t = $row['Time'];
date('Y-m-d\TH:i:s', strtotime($t)); // This got it perfectly

However I still voted it up becauce of the explanation.

Answer 3

$tripid=$_REQUEST['tripid'];
$sql="SELECT * FROM tripdetails WHERE trip_id='".$tripid."'";
$trpresult=mysqli_query($connect,$sql);
  if(mysqli_num_rows($trpresult)==1)
    {
      $trpdetails=mysqli_fetch_assoc($trpresult);
    }
 $trpstartdate = substr_replace($trpdetails['trip_start_date'],T,11,0);

 $string = preg_replace('/\s+/', '', $trpstartdate);

This is Html part

<input type="datetime-local" name="trip_start_date" id="cal" value="<?php echo $string?>">

Answer 4

You can use

date('Y-m-d\TH:i'); //Example result: '2017-01-01T01:01'

if use \T instead of T (not working)

date('Y-m-dTH:i'); //Example result: '2017-01-01UTC01:01'

Answer 5

When submitting <form> using <input type="datetime-local">

the value format you will get is look like this.

2019-09-06T00:21

To set new value in your input type box.

You must use:

date('Y-m-d\TH:i', strtotime($exampleDate)) //2019-08-18T00:00

Solution Example:

$exampleDate = "2019-08-18 00:00:00";//sql timestamp
$exampleDate = strtotime($exampleDate);
$newDate = date('Y-m-d\TH:i', $exampleDate);

or

$exampleDate = "2019-08-18 00:00:00";//sql timestamp
$newDate = date('Y-m-d\TH:i', strtotime($exampleDate));

If you dont use strtotime() you will get an error of

Notice: A non well formed numeric value encountered

---

Tested:

 - $exampleDate = 2019-08-18 00:00:00 ;
 - //Not Working - output(1970-01-01T01:33:39)
 - <?php echo date('Y-m-d\TH:i:s', $exampleDate);?>
 - //Not Working - output(1970-01-01T01:33:39+01:00)
 - <?php echo date('Y-m-d\TH:i:sP', $exampleDate);?>
 - //Not Working - output(2019-08-18T00:00:00+02:00)
 - <?php echo date("c", strtotime($exampleDate));?>
 - //Not Working - output(2019-09-23T19:36:01+02:00)
 - <?php echo (new DateTime($row['Time']))->format('c');?>
 - //Working Perfect - output(2019-08-18T00:00:00)
 - <?php echo date('Y-m-d\TH:i:s', strtotime($exampleDate));?> 

Answer 6

it's simple is that and working for me first convert your php value to this format

 <?php  $datetime = new DateTime($timeinout[0]->time_in);   ?>

then in value of html input element use this format

<input type="datetime-local" id="txt_time_in" placeholder="Time In" name="timein" value = "<?php echo $datetime->format('Y-m-d\TH:i:s'); ?>" class="form-control" /> 

this will set your value to input element