Question

我有日期女巫的格式为d-m-y，如22-10-49我想将其转换为Date对象以将其保存在我的mysql数据库中。我试试：

DateTime::createFromFormat('d-m-y', $mydate , new DateTimeZone('GMT'));

但结果是2049-10-22而不是1949-10-22。我搜索并得知createFromFormat仅在1970年之后返回日期。但我不知道该怎么做。

P.s：22-10-49就是我所拥有的，还有其他几个这样的日期，我无法改变它的格式或将其转换为22-10-1949或任何其他格式。 P.S 2：＆＃34;我正在处理生日并且感谢22-10-15是1915而不是2015年。

Answer 1

尝试此功能。

修改：首先，我们将转换4位数的两位数年份。然后我们将形成完整的日期并将其传递给函数。

 $original_date = '22-10-49';
    $date_part = explode('-',$original_date);

    $baseyear = 1900; // range is 1900-2062
    $shortyear = $date_part[2];
    $year = 100 + $baseyear + ($shortyear - $baseyear) % 100;
    $subdate = substr( $original_date, 0, strrpos( $original_date, '-' ) );
    $string = $subdate."-".$year;

    echo safe_strtotime($string);

function safe_strtotime($string)
{
    if(!preg_match("/\d{4}/", $string, $match)) return null; //year must be in YYYY form
    $year = intval($match[0]);//converting the year to integer
    if($year >= 1970) return date("Y-m-d", strtotime($string));//the year is after 1970 - no problems even for Windows
    if(stristr(PHP_OS, "WIN") && !stristr(PHP_OS, "DARWIN")) //OS seems to be Windows, not Unix nor Mac
    {
        $diff = 1975 - $year;//calculating the difference between 1975 and the year
        $new_year = $year + $diff;//year + diff = new_year will be for sure > 1970
        $new_date = date("Y-m-d", strtotime(str_replace($year, $new_year, $string)));//replacing the year with the new_year, try strtotime, rendering the date
        return str_replace($new_year, $year, $new_date);//returning the date with the correct year
    }
    return date("Y-m-d", strtotime($string));//do normal strtotime
}

输出：1949-10-22

来源：Using strtotime for dates before 1970

以特定格式在PHP中为1970年之前的日期创建Date对象

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