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。
答案 0 :(得分:0)
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');
答案 1 :(得分:0)
使用Y作为年份来设置1954年。
$in = "121654";
$in19 = Substr($in, 0,4) . "19". Substr($in, 4);
$date = DateTime::createFromFormat('mdY',$in19);
echo $date->format('Y-m-d');
答案 2 :(得分:0)
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 $newDate->format('Y-m-d');
答案 3 :(得分:0)
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,我会通过修复数据来接近它:
MMDDYY
您甚至可以在其间添加一个步骤来测试一些自定义截止年份,例如:
list($m, $d, $y) = str_split('121654', 2);
$date = DateTime::createFromFormat('mdY', "$m{$d}19$y");