Symfony DateTimeType 没有秒/当天的 TimeType [英] Symfony DateTimeType without seconds / TimeType for current day

问题描述

对于我的项目，我需要多个具有特殊"配置的 DateTimeType - 但我根本不知道如何实现这些目标(Symfony 5).

For my project I need multiple DateTimeTypes with "special" configuration - but I simply have no idea how to achieve those things (Symfony 5).

1.没有秒的 DateTimeType(但仍然是 HTML5)

 $builder
    ->add('arrival', DateTimeType::class, ['label' => 'Arrival', 'input' => 'datetime_immutable', 'widget' => 'single_text']);

这会呈现一个类型为 datetime-local 的 HTML5 输入字段，这正是我想要的 - 除了它还有几秒钟:

This renders a HTML5 input field with type datetime-local which is exactly what I want - except it still has seconds:

我希望删除 :46.这显然来自 symfony 自动(?)设置的 value 选项:

I want the :46 removed. This clearly comes from the value option which is automatically(?) set by symfony:

从技术上讲，这已经适用于 chrome 等.但是由于秒数，这在 Safari 中无法通过客户端验证.

Technically this already works in chrome etc. However this fails the client-side validation in Safari because of the seconds.

如何去除秒数?format 选项不能使用(因为它需要将 html5 设置为 false)！

How can I remove the seconds? The format option can not be used (because it requires html5 to be set to false)!

2.当天的时间类型

如果您将 TimeType 与 DateTime 对象一起使用，则日期将始终为 1970-01-01，这完全有意义 - 但是您如何预先选择对象的日期(例如今天)并让用户选择时间?

If you use the TimeType with a DateTime object, the date will always be 1970-01-01 which makes total sense - however how can you pre-select the date of the object (e.g. today) and let the user pick the time?

我怎样才能做到这一点?

推荐答案

为了设置今天的默认值，我在实体对象中使用了 construct 方法.

To set the default value for today, I am using the construct method in my entity object.

    public function __construct()
    {
        $this->arrival = new \DateTimeImmutable();
        $this->departure = \new \DateTimeImmutable();
    }

第一季度的答案:为了删除秒，只需创建一个没有秒的新 DateTime 对象.

Answer for Q1: In order to remove the seconds, just create a new DateTime object without the seconds.

    public function __construct()
    {
        $this->arrival = \DateTimeImmutable::createFromFormat('Y-m-d H:i', date('Y-m-d H:i'));
        $this->departure = \DateTimeImmutable::createFromFormat('Y-m-d H:i', date('Y-m-d H:i'));
    }

    //$this->arrival = new \DateTimeImmutable(date('Y-m-d H:i')); //this also works, but needs execption handling (try/catch)

Q2 的答案:此问题的答案相同 - 只需在实体中预先设置当前日期并使用 TimeType 让用户选择时间.

Answer for Q2: Same answer for this question - just pre-set the current date in the entity and use the TimeType to let the user pick the time.

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