我从一个php文件接收到一个日期字段，我需要格式化它。
我得到这个字符串：“2009-11-12 17:58:13”
我想改成“2009年11月12日下午5:58:13”
我试着和约会班合作，但是我得到了这样的东西：
类型强制失败：无法将“2009-11-12 17:58:13”转换为日期。
有人知道做这个有什么好的工具吗？
最佳答案
如果没有直接的解决方案，可以使用regex：
private function formatDate(str:String):String
{
var regex:RegExp = /(\d+)-(\d+)-(\d+)\s(\d+):(\d+):(\d+)/;
var matches:Object = regex.exec(str);
var date:Date = new Date(matches[1], Number(matches[2]) - 1, matches[3],
matches[4], matches[5], matches[6]);
var months:Array = ["Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug",
"Sep", "Oct", "Nov", "Dec"];
var result:String = months[date.month] + " ";
result += date.date + ", " + date.fullYear + " ";
if(date.hours > 12)
result += (date.hours - 12) + ":" + date.minutes + ":" +
date.seconds + " pm";
else
result += date.hours + ":" + date.minutes + ":" + date.seconds + " am";
return result;
}
regex的解释
/(\d+)-(\d+)-(\d+)\s(\d+):(\d+):(\d+)/// Forward slashes - / - at the beginning and end are delimiters
\d+ /* One or more number of digits.
* Ideally it should be \d{n} (\d{4} - 4 digits),
* but I'm assuming that the input string would always be
* valid and properly formatted
*/
- and : //A literal hyphen/colon
\s //A whitespace (use \s+ to skip extra whitespaces)
() /* Parenthetic groups used to capture a string for using it later
* The return value of regex.exec is an array whose first element is
* the complete matched string and subsequent elements are contents
* of the parenthetic groups in the order they appear in the regex
*/
关于php - as3-从mysql格式化时间戳字符串，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/1726586/