我一直试图找到一种非常快速的方法来将yyyy-mm-dd [hh:mm:ss]解析为Date对象。以下是我尝试过的3种方法以及每种方法解析50,000个日期时间字符串所需的时间。
有没有人知道更快的方法或提示加快方法?
castMethod1 takes 3673 ms
castMethod2 takes 3812 ms
castMethod3 takes 3931 ms
代码:
private function castMethod1(dateString:String):Date {
if ( dateString == null ) {
return null;
}
var year:int = int(dateString.substr(0,4));
var month:int = int(dateString.substr(5,2))-1;
var day:int = int(dateString.substr(8,2));
if ( year == 0 && month == 0 && day == 0 ) {
return null;
}
if ( dateString.length == 10 ) {
return new Date(year, month, day);
}
var hour:int = int(dateString.substr(11,2));
var minute:int = int(dateString.substr(14,2));
var second:int = int(dateString.substr(17,2));
return new Date(year, month, day, hour, minute, second);
}
-
private function castMethod2(dateString:String):Date {
if ( dateString == null ) {
return null;
}
if ( dateString.indexOf("0000-00-00") != -1 ) {
return null;
}
dateString = dateString.split("-").join("/");
return new Date(Date.parse( dateString ));
}
-
private function castMethod3(dateString:String):Date {
if ( dateString == null ) {
return null;
}
var mainParts:Array = dateString.split(" ");
var dateParts:Array = mainParts[0].split("-");
if ( Number(dateParts[0])+Number(dateParts[1])+Number(dateParts[2]) == 0 ) {
return null;
}
return new Date( Date.parse( dateParts.join("/")+(mainParts[1]?" "+mainParts[1]:" ") ) );
}
不,默认情况下,Date.parse不会处理破折号。我需要为"0000-00-00"
等日期时间字符串返回null。
答案 0 :(得分:17)
我一直在使用以下snipplet来解析UTC日期字符串:
private function parseUTCDate( str : String ) : Date {
var matches : Array = str.match(/(\d\d\d\d)-(\d\d)-(\d\d) (\d\d):(\d\d):(\d\d)Z/);
var d : Date = new Date();
d.setUTCFullYear(int(matches[1]), int(matches[2]) - 1, int(matches[3]));
d.setUTCHours(int(matches[4]), int(matches[5]), int(matches[6]), 0);
return d;
}
只需删除时间部分,它应该可以满足您的需求:
private function parseDate( str : String ) : Date {
var matches : Array = str.match(/(\d\d\d\d)-(\d\d)-(\d\d)/);
var d : Date = new Date();
d.setUTCFullYear(int(matches[1]), int(matches[2]) - 1, int(matches[3]));
return d;
}
不知道速度,我在我的应用程序中并没有担心。在我的机器上显着不到一秒的50K次迭代。
答案 1 :(得分:5)
这是一些摆弄后我能想到的最快的:
private function castMethod4(dateString:String):Date {
if ( dateString == null )
return null;
if ( dateString.length != 10 && dateString.length != 19)
return null;
dateString = dateString.replace("-", "/");
dateString = dateString.replace("-", "/");
return new Date(Date.parse( dateString ));
}
我在计算机上为castMethod2()获得了大约470ms的50k迭代,对于我的版本获得了300毫秒(这与63%的时间内完成的工作量相同)。我肯定会说两者都“足够好”,除非你解析愚蠢的日期。
答案 2 :(得分:2)
我猜测Date.Parse()不起作用?
答案 3 :(得分:1)
那么方法2似乎是最好的方法:
private function castMethod2(dateString:String):Date {
if ( dateString == null ) {
return null;
}
if ( dateString.indexOf("0000-00-00") != -1 ) {
return null;
}
dateString = dateString.split("-").join("/");
return new Date(Date.parse( dateString ));
}
答案 4 :(得分:1)
因为Date.parse()不接受所有可能的格式,我们可以使用DateFormatter和Data.parse()可以理解的formatString预先格式化传递的dateString值,例如
// English formatter
var stringValue = "2010.10.06"
var dateCommonFormatter : DateFormatter = new DateFormatter();
dateCommonFormatter.formatString = "YYYY/MM/DD";
var formattedStringValue : String = dateCommonFormatter.format(stringValue);
var dateFromString : Date = new Date(Date.parse(formattedStringValue));
答案 5 :(得分:1)
var strDate:String = "2013-01-24 01:02:40";
function dateParser(s:String):Date{
var regexp:RegExp = /(\d{4})\-(\d{1,2})\-(\d{1,2}) (\d{2})\:(\d{2})\:(\d{2})/;
var _result:Object = regexp.exec(s);
return new Date(
parseInt(_result[1]),
parseInt(_result[2])-1,
parseInt(_result[3]),
parseInt(_result[4]),
parseInt(_result[5]),
parseInt(_result[6])
);
}
var myDate:Date = dateParser(strDate);
答案 6 :(得分:0)
这是我的实施。试一试。
public static function dateToUtcTime(date:Date):String {
var tmp:Array = new Array();
var char:String;
var output:String = '';
// create format YYMMDDhhmmssZ
// ensure 2 digits are used for each format entry, so 0x00 suffuxed at each byte
tmp.push(date.secondsUTC);
tmp.push(date.minutesUTC);
tmp.push(date.hoursUTC);
tmp.push(date.getUTCDate());
tmp.push(date.getUTCMonth() + 1); // months 0-11
tmp.push(date.getUTCFullYear() % 100);
for(var i:int=0; i < 6/* 7 items pushed*/; ++i) {
char = String(tmp.pop());
trace("char: " + char);
if(char.length < 2)
output += "0";
output += char;
}
output += 'Z';
return output;
}