我个人更喜欢使用高级对象而不仅仅是数组或字符串，所以这是我处理的方式。对于有些人来说，这可能有些过度，我完全理解，不过核心方法也可以很容易地重写为全局函数，希望这一点可以考虑到。

我认为整体上非常简单明了，几乎与您的代码完全相同，只是在分割后我们将其转换为整数，然后验证小时是否在预期范围内。

readonly class MySqlTime
{
    public function __construct(
        public int $hours,
        public int $minutes,
        public int $seconds,
    ) {
    }

    public static function fromMySqlTimeAsDurationInDay(string $time): self
    {
        [$hours, $minutes, $seconds] = array_map('intval', explode(':', $time));

        // Ensure only positive values on a single day.
        // MySql should already be throwing errors about minutes or seconds being out of range, so we don't
        // need to check those.
        // The seconds could include decimals, however for this purpose we are ignoring them.
        if ($hours > 23 || $hours < 0) {
            throw new InvalidArgumentException('Hours must be between 0 and 23');
        }

        return new self($hours, $minutes, $seconds);
    }

    public function toDateTime(DateTimeInterface $dateTime = null) : DateTimeInterface
    {
        $dateTime ??= new DateTimeImmutable();

        return $dateTime->setTime($this->hours, $this->minutes, $this->seconds);
    }
}

用法：

var_dump(MySqlTime::fromMySqlTimeAsDurationInDay('12:34:56')->toDateTime());

输出：

object(DateTimeImmutable)#3 (3) {
  ["date"]=>
  string(26) "2023-07-31 12:34:56.000000"
  ["timezone_type"]=>
  int(3)
  ["timezone"]=>
  string(16) "Europe/Amsterdam"
}

Demo: https://3v4l.org/B6mr4#v8.2.7