php 如何为输入类型“datetime-local”设置值？

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

提问by Jbadminton

I tried this:

我试过这个：

<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!

如何使用数据库中的数据设置此输入字段的值？
它不起作用！！
我也需要让编辑成为可能。
或者我应该使用其他类型的输入？
$row['Time']来自数据库！

回答by Karol Gasienica

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

我不知道具体是什么，$row['Time']但它应该如下：

Definition

定义

A valid date-time as defined in RFC 3339with 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

RFC 3339 中定义的有效日期时间，具有以下附加条件：

• 日期/时间语法中的文字字母 T 和 Z 必须始终大写
• date-fullyear 生产被定义为代表大于 0 的数字的四位或更多位数字

例子

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

---

---

Solution

解决方案

To create RFC 3339format in PHP you can use:

要在 PHP 中创建RFC 3339格式，您可以使用：

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

or in another way:

或以另一种方式：

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

or if you prefer objective style:

或者如果您更喜欢客观风格：

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

---

---

In your code

在你的代码中

So in your code it would look as follows:

因此，在您的代码中，它将如下所示：

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

or

或者

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

---

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Manual

手动的

• More informations can be found here

• PHP date Manual

• PHP DateTime Manual

• 更多信息可以在这里找到

• PHP日期手册

• PHP 日期时间手册

回答by Mohsin Shoukat

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

很简单，为我工作首先将您的 php 值转换为这种格式

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

then in value of html input element use this format

然后在 html input 元素的值中使用这种格式

<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

这会将您的值设置为输入元素

回答by Ferhat KO?ER

You can use

您可以使用

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

if use \T instead of T (not working)

如果使用 \T 而不是 T （不工作）

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

回答by ailia

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

Karol Gasienica 的回答是一个很好的解释，但不知何故，即使在他们的回复中也对我不起作用

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

对我有用的是这个

$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.

但是，由于解释，我仍然投票赞成。

回答by Rhalp Darren Cabrera

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

提交时<form>使用<input type="datetime-local">

the value format you will get is look like this.

您将获得的值格式如下所示。

2019-09-06T00:21

2019-09-06T00:21

To set new value in your input type box.

在输入类型框中设置新值。

You must use:

您必须使用：

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

Solution Example:

解决方案示例：

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

or

或者

$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

如果你不使用strtotime()你会得到一个错误

Notice: A non well formed numeric value encountered

注意：遇到格式不正确的数值

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Tested测试：

 - $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by Deepak Sharma

This will convert datetime from database to datetime-local

这会将日期时间从数据库转换为日期时间本地

str_replace(" ","T",substr_replace($string ,"", -3))

str_replace(" ","T",substr_replace($string ,"", -3))

回答by danish-khan-I

None of the above solutions worked for me as of 2019 using Google Chrome Version 78.0.3904.70 (Official Build) (64-bit)

截至 2019 年，使用 Google Chrome 的上述解决方案均不适用于我 Version 78.0.3904.70 (Official Build) (64-bit)

What worked for me is.

对我有用的是。

<input type="datetime-local" value="2017-06-13T13:00">

As you can see the format is 2017-06-13T13:00or Y-m-dTH:i.

如您所见，格式为2017-06-13T13:00或Y-m-dTH:i。

As of PHP you can do like.

从 PHP 开始，您可以这样做。

<input type="datetime-local" value="<?php echo Date('Y-m-d\TH:i',time()) ?>">

Hope this will save someone's time. :)

希望这会节省某人的时间。:)

回答by Rishi

$tripid=$_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

这是 Html 部分

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