问题
I tried this:
<input type="datetime-local" value="<?php echo $row['Time']; ?>" class="date" name="start" REQUIRED>
How can I set the value of this input field with the data from the database?
It doesn't work!!
I need to make it possible to edit too.
Or should I use another type of input?$row['Time'] is from the database!
回答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
回答2:
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 $datetime->format('Y-m-d\TH:i:s'); ?>" class="form-control" />
this will set your value to input element
回答3:
You can use
date('Y-m-d\TH:i'); //Example result: '2017-01-01T01:01'
if use \T instead of T (not working)
date('Y-m-dTH:i'); //Example result: '2017-01-01UTC01:01'
回答4:
When submitting <form> using <input type="datetime-local">
the value format you will get is look like this.
2019-09-06T00:21
To set new value in your input type box.
You must use:
date('Y-m-d\TH:i', strtotime($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));?>
回答5:
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.
回答6:
$tripid=$_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 $string?>">
回答7:
This will convert datetime from database to datetime-local
str_replace(" ","T",substr_replace($string ,"", -3))
回答8:
None of the above solutions worked for me as of 2019 using Google Chrome Version 78.0.3904.70 (Official Build) (64-bit)
What worked for me is.
<input type="datetime-local" value="2017-06-13T13:00">
As you can see the format is 2017-06-13T13:00 or Y-m-dTH:i.
As of PHP you can do like.
<input type="datetime-local" value="<?php echo Date('Y-m-d\TH:i',time()) ?>">
Hope this will save someone's time. :)
来源：https://stackoverflow.com/questions/40843407/how-to-set-a-value-for-the-input-type-datetime-local