格式化日期时间 [英] Format DateTime

问题描述

我正在尝试格式化当前的 datetime ，然后使用休眠将其保存到数据库中.我将其保存为默认日期格式，即:

I'm trying to format the current datetime and then save it to a database using hibernate. I got it to save using the default date format, i.e.:

Date date = new Date();
hibernateObject.setDate(date);
hibernateObject.save(date);

我找到了一种格式化它的方法，但是它确实很不整洁，而且似乎不起作用:

I've found a way to format it but it's really untidy and doesn't seem to work:

Date finalDate = null;
try {
    Date date = new Date();
    DateFormat df = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss");
    String sDate = df.format(date);
    finalDate = df.parse(sDate);
} catch (ParseException pe) {
    System.out.println("ParserException while attempting to establish date");
}
hibernateObject.setDate(date);
hibernateObject.save(date);

推荐答案

尝试使用此方法(在您的hibernateObject类中):

Try to use this (in your hibernateObject class):

@Temporal(TemporalType.TIMESTAMP)
@DateTimeFormat(pattern = "yyyy-MM-dd hh:mm:ss")
public Date getStartDate() {
    return startDate;
}
public void setStartDate(Date startDate) {
    this.startDate = startDate;
}

当您执行显示操作时，日期对象实际上没有任何更改.唯一发生的事情:您创建包含格式化日期的字符串.因此，您可以指定DateTimeFormat模式以将日期存储在该模式中.

When you do as you show date object actually has no changes. Only thing that happens: you create String that contains formatted date. So you can specify DateTimeFormat pattern to store date in that pattern instead.

还请注意，日期模式中的"hh"字符串实际上将为您提供1到12之间的值.这是AM/PM小时.要获取0-24小时的值，请改用"HH"(请参见 http://docs.oracle.com/javase/6/docs/api/java/text/SimpleDateFormat.html ，日期和时间模式"部分)

Also note, that "hh" string in your date pattern actually will give you value from 1 to 12. It is AM/PM hour. To get 0-24 hours value use "HH" instead (see http://docs.oracle.com/javase/6/docs/api/java/text/SimpleDateFormat.html, "Date and Time Patterns" section)

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