Question

我正在格式化像'121654'这样的日期，如下所示;

$date = DateTime::createFromFormat('mdy','121654');
echo $date->format('Y-m-d');

现在输出是;

2054年12月16日

但我希望它像1954-12-16。

直到1970年，它给出了正确的结果。但如果年份低于1970年，它将成为未来的日期。

有没有办法按我的意愿制作东西。

Answer 1

您应该使用Y代替y并在此处指定1954。

$date = DateTime::createFromFormat('mdY','12161954');
echo $date->format('Y-m-d');

Answer 2

使用Y作为年份来设置1954年。

https://3v4l.org/4K7Ti

$in = "121654";
$in19 = Substr($in, 0,4) . "19". Substr($in, 4);

$date = DateTime::createFromFormat('mdY',$in19);
echo $date->format('Y-m-d');

Answer 3

您可以检查生成的日期是否小于1970年，然后您可以使用函数将 X 年添加到其中，因此将来会是这样：

$date = DateTime::createFromFormat('mdY','12161960');
$date1970 = DateTime::createFromFormat('mdY','01011970');
if ($date < $date1970) {
    $newDate = $date->modify('+500 years');
}
echo $newDate->format('Y-m-d');

Answer 4

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MMDDYY

您甚至可以在其间添加一个步骤来测试一些自定义截止年份，例如：

list($m, $d, $y) = str_split('121654', 2);
$date = DateTime::createFromFormat('mdY', "$m{$d}19$y");

PHP日期格式将一年格式化为未来日期（如2050年），而不是1950年日期小于1970

4 个答案: