我使用转换器（这里的源代码：https://github.com/lingtalfi/DatePickerHelper/blob/master/DatePickerHelper.php）
用法：
<?php
echo DatePickerHelper::convertFromDatePickerToPhpDate("dd/mm/yy"); // prints d/m/Y
echo DatePickerHelper::convertFromPhpDateToDatePicker("Y-m-d"); // prints yy-mm-dd
$birthdayDate = "09/12/1944"; // your input from the web
echo DatePickerHelper::convertFromNumericInputToMysqlDate($birthdayDate, "d/m/Y"); // 1944-12-09
-
class DatePickerHelper
{
private static $map = [
"yy" => "<1>",
"y" => "<2>",
"MM" => "<3>",
"M" => "<4>",
"mm" => "<5>",
"m" => "<6>",
"DD" => "<7>",
"D" => "<8>",
"oo" => "<9>",
"o" => "<10>",
"dd" => "<11>",
"d" => "<12>",
];
private static $map2 = [
"<1>" => 'Y',
"<2>" => 'y',
"<3>" => 'F',
"<4>" => 'M',
"<5>" => 'm',
"<6>" => 'n',
"<7>" => 'l',
"<8>" => 'D',
"<9>" => 'z', // note: php doesn't have "day of the year with three digits", but this is the closest
"<10>" => 'z',
"<11>" => 'd',
"<12>" => 'j',
];
private static $mapRegex = [
'Y' => '<1>',
'y' => '<2>',
'm' => '<3>',
'n' => '<4>',
'd' => '<5>',
'j' => '<6>',
];
private static $mapRegex2 = [
'<1>' => '(?P<year4>[0-9]{4})',
'<2>' => '(?P<year2>[0-9]{2})',
'<3>' => '(?P<month_leading>[0-9]{2})',
'<4>' => '(?P<month_no_leading>[0-9]{1,2})',
'<5>' => '(?P<day_leading>[0-9]{2})',
'<6>' => '(?P<day_no_leading>[0-9]{1,2})',
];
public static function convertFromDatePickerToPhpDate(string $datePickerFormat): string
{
$map = self::$map;
$map2 = self::$map2;
$first = str_replace(array_keys($map), array_values($map), $datePickerFormat);
return str_replace(array_keys($map2), array_values($map2), $first);
}
public static function convertFromPhpDateToDatePicker(string $phpDate): string
{
$map2 = array_flip(self::$map2);
$map = array_flip(self::$map);
$first = str_replace(array_keys($map2), array_values($map2), $phpDate);
return str_replace(array_keys($map), array_values($map), $first);
}
/**
* @param string $input , the string to convert, the format of this string should match the given phpFormat
* Plus, it must contain exactly:
* - one day component
* - one month component
* - one year component
*
* @param string $phpFormat , all components of the phpFormat have to be one of those:
* - Y: year, four digits
* - y: year, two digits
* - m: numeric month, with leading zeros
* - n: numeric month, without leading zeros
* - d: numeric day of the month, with leading zeros
* - j: numeric day of the month, without leading zeros
*/
public static function convertFromNumericInputToMysqlDate(string $input, string $phpFormat)
{
$map = self::$mapRegex;
$map2 = self::$mapRegex2;
$first = str_replace(array_keys($map), array_values($map), $phpFormat);
$pattern = str_replace(array_keys($map2), array_values($map2), $first);
if (preg_match('!' . $pattern . '!', $input, $match)) {
$day = $match['day_leading'] ?? $match['day_no_leading'] ?? null;
if (null !== $day) {
$day = (int)$day;
$month = $match['month_leading'] ?? $match['month_no_leading'] ?? null;
if (null !== $month) {
if (
array_key_exists("year4", $match) ||
array_key_exists("year2", $match)
) {
// a component of each type is there, we will be able to return a result
if (array_key_exists("year4", $match)) {
$year = (int)$match['year4'];
} else {
// assumed it's 20, but we don't know really, that sucks.
// That's why you should use year4 instead...
$year = "20" . $match['year2'];
$year = (int)$year;
}
return $year . "-" . sprintf('%02s', $month) . "-" . sprintf("%02s", $day);
}
}
}
}
return false;
}
}