了解JodaTime DateTime.parse（string，formatter） [英] Understanding JodaTime DateTime.parse(string, formatter)

问题描述

格式化程序在解析

Does the style of the formatter in the parse method of the DateTime class have to match the exact style of the string? For instance, I'm getting a TimeStamp object from the database (Oracle) and converting it to a string. In the database the TimeStamp is stored like this

08-AUG-12 12.00.00.000000000 AM

08-AUG-12 12.00.00.000000000 AM

我将格式化器设置为此样式

I set my formatter to this style

String pattern = "dd-MMM-yy";

我得到这个例外

java.lang.IllegalArgumentException: Invalid format: "08-AUG-12 12.00.00 AM" is malformed at " 12.00.00 AM"

org.joda.time.format.DateTimeFormatter.parseDateTime(DateTimeFormatter.java:866)

org.joda.time.DateTime.parse(DateTime.java:144)

这是什么意思，我该如何解决？当我将格式化程序设置为yy-MMM-dd hh.mm.ss aa我没有得到例外，但它在浏览器中打印如下： 2008-08-12T00：00：00.000-04：00 ，但是我需要它作为打印出来dd-MMM-yy hh：mm： ss aa

What exactly does this mean and how would I go about fixing it? When I set my formatter to "yy-MMM-dd hh.mm.ss aa" I don't get an exception but it prints in the browser like this: 2008-08-12T00:00:00.000-04:00, but I need for it to print out as "dd-MMM-yy hh:mm:ss aa"

推荐答案

使用LocalDateTime：

Use LocalDateTime instead:

String input = "08-AUG-12 12.00.00 AM";
String pattern = "dd-MMM-yy hh.mm.ss aa";

LocalDateTime localDateTime = LocalDateTime.parse(input, DateTimeFormat.forPattern(pattern));

编辑

事实上你可以使用DateTime来执行它：

As a matter of fact you can do it with DateTime also:

private static String parseDateTime(String input){
     String pattern = "dd-MMM-yy hh.mm.ss aa";
     DateTime dateTime  = DateTime.parse(input, DateTimeFormat.forPattern(pattern));
     return dateTime.toString("dd-MMM-yy hh:mm:ss aa");
}

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