我有一个用户格式的日期和使用PHP日期格式(http://fr.php.net/manual/en/function.date.php)的对应格式。用户输入前不知道此格式。
我必须将此日期转换为已知格式,例如'd / m / Y h:i:s'。 我的PHP服务器使用5.2.1 PHP版本,所以我不能使用DateTime类。
有人可以选择DateTime :: createFromFormat方法,这样做的脚本或想法可以帮助我吗?
祝你好运, 弗洛里安。
修改:
我编写了一个函数,用于将日期转换为其他格式而不知道原始日期格式:
/**
* This function converts a date with unknown format into a defined format.
* It could detect some standard date formats : dd/mm/YYYY, dd-mm-YY, YYYY-mm-dd hh:mm:ss, etc.
* @param string $date date which is user formatted.
* @param int $options used to help algorithm to detect date format : US or FR date, past date or future date, hour format.
* @param string $newFormat format the date must be converted into
* @param string $originalFormat format detected
* @return boolean|string the new date, corresponding to the $date argument, formatted into the $newFormat format
*/
function convertDateToFormat($date, $options = 0, $newFormat, &$originalFormat = ''){
if($options == 0){
$options = DATE_FR | DATE_PAST | HOUR_24;
}
$matches = array();
$year = 0;
$month = 0;
$day = 0;
$hour = 0;
$minute = 0;
$second = 0;
$originalFormatOk = false;
if(preg_match('#(\d{1,2})([/.-])(\d{1,2})(([/.-])(\d{2,4}))?#', $date, $matches)){
if(isset($matches[6])){
if(strlen($matches[6]) == 2){
if($matches[6] >= date('y') && DATE_PAST)
$year = '19'.$matches[6];
else
$year = '20'.$matches[6];
$originalFormatyear = 'y';
if($options & DATE_US){
$originalFormat = 'm'.$matches[2].'d'.$matches[5].$originalFormatyear;
$year = $matches[6];
$month = $matches[1];
$day = $matches[3];
$originalFormatOk = true;
}
}
else{
$year = $matches[6];
$originalFormatyear = 'Y';
}
if(!$originalFormatOk){
$year = $matches[6];
$month = $matches[3];
$day = $matches[1];
$originalFormat = 'd'.$matches[2].'m'.$matches[5].$originalFormatyear;
$originalFormatOk = true;
}
}
else{
if($options & DATE_US){
$originalFormat = 'm'.$matches[2].'d';
$year = date('Y');
$month = $matches[1];
$day = $matches[3];
}
else{
$originalFormat = 'd'.$matches[2].'m';
$year = date('Y');
$month = $matches[3];
$day = $matches[1];
}
$originalFormatOk = true;
}
}
if(preg_match('#'.$matches[0].'(\D*)(\d{1,2})([:hH])(\d{1,2})(:(\d{1,2}))?#', $dateHour, $matches)){
if($options & HOUR_12)
$originalFormatHour = 'h';
else
$originalFormatHour = 'H';
$hour = $matches[2];
$minute = $matches[4];
if(strtolower($matches[3]) == 'h')
$matches[3] = '\\'.$matches[3];
$originalFormat .= $matches[1].$originalFormatHour.$matches[3].'i';
if(isset($matches[6])){
$second = $matches[6];
$originalFormat .= ':s';
}
}
if($originalFormatOk)
return date($newFormat, mktime($hour, $minute, $second, $month, $day, $year));
return false;
}
答案 0 :(得分:2)
date_parse_from_format()
功能可以完成您需要的一切。
E.g。 (来自文档):
$date = "6.1.2009 13:00+01:00";
print_r(date_parse_from_format("j.n.Y H:iP", $date));
但是,如果你不知道格式,那或多或少粗略猜测你必须使用类似date_parse()
的东西,它会给你(希望)一个正确的日期但是可能会失败。
E.g。 (来自文档):
print_r(date_parse("2006-12-12 10:00:00.5"));
将打印:
Array
(
[year] => 2006
[month] => 12
[day] => 12
[hour] => 10
[minute] => 0
[second] => 0
[fraction] => 0.5
[warning_count] => 0
[warnings] => Array()
[error_count] => 0
[errors] => Array()
[is_localtime] =>
)